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BARRETT O'NEILL
ELEME NTARY
DIFFERE NTIAL
GEOMETRY
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Elementary
Differential Geometry
Elementary
Differential Geometry
Barrett O'Neill
DEPARTMENT OF MATHEMATICS
UNIVERSITY OF CALIFORNIA
LOS ANGELES, CALIFORNIA
ACADEMIC PRESS New York San Francisco London
A Subsidiary of Harcourt Brace Jovanovich, Publishers
Copyright © 1966, by Academic Press, Inc.
all rights reserved.
no part of this publication may be reproduced or
transmitted in any form or by any means, electronic
or mechanical, including photocopy, recording, or any
information storage and retrieval system, without
permission in writing from the eublisher.
ACADEMIC PRESS, INC.
Ill Fifth Avenue, New York, New York 10003
United Kingdom Edition published by
ACADEMIC PRESS, INC. (LONDON) LTD.
24/28 Oval Road, London NW1
Library of Congress Catalog Card Number: 6614468
PRINTED IN THE UNITED STATES OF AMERICA
Preface
This book is an elementary account of the geometry of curves and surfaces.
It is written for students who have completed standard first courses in
calculus and linear algebra, and its aim is to introduce some of the main
ideas of differential geometry.
The traditional undergraduate course in differential geometry has changed
very little in the last few decades. By contrast, geometry has been ad
vancing very rapidly at the research level, and there is general agreement
that the undergraduate course needs to be brought up to date. I have tried
to think through the classical material, to prune and augment it, and to
write down the results in a reasonably clean and modern mathematical style
However, I have used a new idea only if it really pays its way by simplify
ing and clarifying the exposition.
Chapter I establishes the language of the book — a language compounded of
familiar parts of calculus and linear algebra. Chapter II describes the
method of "moving frames," which is introduced, as in elementary calculus,
to study curves in space. In Chapter III we investigate the rigid motions
of space, in terms of which congruence of curves (or surfaces) in space is
defined in the same fashion as congruence of triangles in the plane.
Chapter IV requires special comment. The main weakness of classical
differential geometry was its lack of any adequate definition of surface.
In this chapter we decide just what a surface is, and show that each surface
has a differential and integral calculus of its own, strictly comparable with
the familiar calculus of the plane. This exposition provides an introduction
to the notion of differ entiable manifold, which has become indispensable to
those branches of mathematics and its applications based on the calculus.
The next two chapters are devoted to the geometry of surfaces in 3space.
Chapter V stresses intuitive and computational aspects to give geometrical
meaning to the theory presented in Chapter VI. In the final chapter, al
though our methods are unchanged, there is a radical shift of viewpoint.
Roughly speaking, we study the geometry of a surface as seen by its inhabi
vi PREFACE
tants, with no assumption that the surface is to be found in ordinary three
dimensional space.
No branch of mathematics makes a more direct appeal to the intuition
than geometry. I have sought to emphasize this by a large number of illus
trations, which form an integral part of the text. A set of exercises appears
at the end of each section; these range from routine tests of comprehension
to more seriously challenging problems.
In teaching from preliminary versions of this book, I have usually cov
ered the background material in Chapter I rather rapidly, and have not
devoted any classroom time to Chapter III (hence also Section 8 of Chapter
VI). A course in the geometry of curves and surfaces in space might consist
of: Chapter II, Chapter IV (omit Sections 6 and 8), Chapter V, and Chapter
VI (omit Sections 6 and 7). This is essentially the content of the tradi
tional undergraduate course in differential geometry, with clarification of
the concepts of surface and mapping of surfaces.
The omitted sections in the list above are used only in Chapter VII. This
final chapter, an extensive account of twodimensional Riemannian geom
etry, is in a sense the goal of the book. Rather than shift the discourse to
higher dimensions, I have preferred to retain dimension 2, so that this more
sophisticated view of geometry will develop directly from the special case
of surfaces in 3space. Chapter VII is long, and on a first reading Theorem
5.9 and Sections 6 and 7 may well be omitted. Serious use of differential
equations theory has been largely avoided in the early chapters; however,
some acquaintance with the fundamentals of the subject will be helpful
in Chapter VII.
Los Angeles, California B. O'N.
Contents
Preface
Introduction l
Chapter I. Calculus on Euclidean Space
1. Euclidean Space 3
2. Tangent Vectors 6
3. Directional Derivatives 11
4. Curves in E 3 15
5. 1Forms 22
6. Differential Forms 26
7. Mappings 32
8. Summary 41
Chapter II. Frame Fields
1. Dot Product 42
2. Curves 51
3. The Frenet Formulas 56
4. ArbitrarySpeed Curves 66
5. Covariant Derivatives 77
6. Frame Fields 81
7. Connection Forms 85
8. The Structural Equations 91
9. Summary 96
Chapter III. Euclidean Geometry
1. Isometries of E 3 98
2. The Derivative Map of an Isometry 104
3. Orientation 107
4. Euclidean Geometry 112
5. Congruence of Curves 116
6. Summary 123
vin CONTENTS
Chapter IV. Calculus on a Surface
1. Surfaces in E 3 124
2. Patch Computations 133
3. Differentiable Functions and Tangent Vectors 143
4. Differential Forms on a Surface 152
5. Mappings of Surfaces 158
6. Integration of Forms 167
7. Topological Properties of Surfaces 176
8. Manifolds 182
9. Summary 187
Chapter V. Shape Operators
1. The Shape Operator of M C E 3 189
2. Normal Curvature 195
3. Gaussian Curvature 203
4. Computational Techniques 210
5. Special Curves in a Surface 223
6. Surfaces of Revolution 234
7. Summary 244
Chapter VI. Geometry of Surfaces in E 3
1. The Fundamental Equations 245
2. Form Computations 251
3. Some Global Theorems 256
4. Isometries and Local Isometries 263
5. Intrinsic Geometry of Surfaces in E 3 271
6. Orthogonal Coordinates 276
7. Integration and Orientation 280
8. Congruence of Surfaces 297
9. Summary 303
Chapter VII. Riemannian Geometry
1. Geometric Surfaces 304
2. Gaussian Curvature 310
3. Covariant Derivative 318
4. Geodesies 326
5. LengthMinimizing Properties of Geodesies 339
6. Curvature and Conjugate Points 352
7. Mappings that Preserve Inner Products 362
8. The GaussBonnet Theorem 372
9. Summary 389
Bibliography 391
Answers to OddNumbered Exercises 393
Index 405
Elementary
Differential Geometry
Introduction
This book presupposes a reasonable knowledge of elementary calculus and
linear algebra. It is a working knowledge of the fundamentals that is
actually required. The reader will, for example, frequently be called upon
to use the chain rule for differentiation, but its proof need not concern us.
Calculus deals mostly with realvalued functions of one or more varia
bles, linear algebra with functions (linear transformations) from one
vector space to another. We shall need functions of these and other types,
so we give here general definitions which cover all types.
A set S is a collection of objects which are called the elements of S. A
set A is a subset of S provided each element of A is also an element of S.
A function f from a set D to a set R is a rule that assigns to each element
x of D a, unique element f(x) of R. The element f(x) is called the value
of / at x. The set D is called the domain of /; the set R is often called the
range of /. If we wish to emphasize the domain and range of a function /,
the notation f:D*R is used. Note that the function is denoted by a single
letter, say /, while fix) is merely a value of /.
Many different terms are used for functions — mappings, transformations,
correspondences, operators, and so on. A function can be described in
various ways, the simplest case being an explicit formula such as
f(x) = 3x 2 + 1,
which we may also write as x — > Zx + 1 .
If both fi and f% are functions from D to R, then fr = / 2 means that
fa( x ) = f 2 ( x ) for all x in D. This is not a definition, but a logical consequence
of the definition of function. JJ
Let f: D —> R and g: E — > S be functions. In general, the image oiyf is
the subset of R consisting of all elements of the form f(x); it is usually
denoted by /(D). Now if this image also happens to be a subset of the
domain E of g, it is possible to combine these two functions to obtain the
composite function g(J): D *• S. By definition, g(f) is the function whose
value on each element x of D is the element g(f(x)) of S.
INTRODUCTION
3
If /: D — > R is a function and A is a subset of D, then the restriction of
/ to A is the function/  A: A — ► R denned by the same rule as/, but applied
only to elements of A. This seems a rather minor change, but the function
/  A may have properties quite different from / itself.
Here are two vital properties which a function may possess. A function
f:D^>R is onetoone, provided that if x and y are any elements of D such
that x 9* y, then f(x) 9* f(y). A function f.D^R is onto (or carries D
onto R) provided that for every element y of R there is at least one element
x of D such that /Or) = y. In short, the image of^fs the entire set R. For
example, consider the following functions, each of which has the real
numbers as both domain and range :
(1 ) The function x — ► x 3 is both onetoone and onto.
(2) The exponential function x — > e x is onetoone, but not onto.
(3) The function x — » x z + x 2 is onto, but not onetoone.
(4) The sine function x — > sin x is neither onetoone nor onto.
If a function /: D — > R is both onetoone and onto, then for each ele
ment y of R there is one and only one element x such that f(x) = y.By
defining/ 1 (y) = x for all x and y so related, we obtain a function/ 1 : R+D
called the inverse of/. Note that the function/ 1 is also onetoone and onto,
and that its inverse function is the original function /.
Here is a short list of the main notations used throughout the book, in
order of their appearance in Chapter I :
P> <I> points (Sec. 1 )
/» 9> realvalued functions (Sec. 1 )
v » w » tangent vectors (Sec. 2)
V>W, vector fields (Sec. 2)
«> Pi curves (Sec. 4)
<k $> differential forms (Sec. 5)
F > G > mappings (Sec. 7)
In Chapter I we define these concepts for Euclidean 3 space. (Extension
to arbitrary dimensions is virtually automatic.) In Chapter IV we show
how these concepts can be adapted to a surface.
A few references are given to the brief bibliography at the end of the
book; these are indicated by numbers in square brackets.
CHAPTER
I
Calculus on Euclidean Space
As mentioned in the Preface, the purpose of this initial chapter is to
establish the mathematical language used throughout the book. Much
of what we do is simply a review of that part of elementary calculus dealing
with differentiation of functions of three variables, and with curves in space.
Our definitions have been formulated so that they will apply smoothly
to the later study of surfaces.
1 Euclidean Space
Threedimensional space is often used in mathematics without being
formally defined. It is said to be the space of ordinary experience. Looking
at the corner of a room, one can picture the familiar process by which
rectangular coordinate axes are introduced and three numbers are meas
ured to describe the position of each point. A precise definition which
realizes this intuitive picture may be obtained by this device: instead of
saying that three numbers describe the position of a point, we define them
to be a point.
1.1 Definition Euclidean Sspace E 3 is the set of all ordered triples of
real numbers. Such a triple p = (pi, p 2 , Pz) is called a point of E 3 .
In linear algebra, it is shown that E 3 is, in a natural way, a vector space
over the real numbers. In fact, if p = (p\,P2,Pz) and q = {q\,qi,qz)
are points of E 3 , their sum is the point
P + q = (pi + gi, pt + g2, ps + gs).
The scalar product of a point p = (p u p 2 , Pz) by a number a is the point
ap = (api, ap 2 , ap z ).
* CALCULUS ON EUCLIDEAN SPACE [Chap. I
It is easy to check that these two operations satisfy the axioms for a vector
space. The point = (0, 0, 0) is called the origin of E 3 .
Differential calculus deals with another aspect of E 3 starting with the
notion of differentiable realvalued functions on E 3 . We recall some funda
mentals.
1 .2 Definition Let x, y, and z be the realvalued functions on E 3 such
that for each point p = (pi,p 2 ,ps)
z(p) = pi, y(p) = Pi, z(p) = p s .
These functions x, y, z are called the natural coordinate functions of E 3 .
We shall also use index notation for these functions, writing
xi = x, x 2 = y, x z = z.
Thus the value of the function Xi on a point p is the number p { , and so
we have tlwydentity p = (pi,p 2 ,p 3 ) = (:ci(p), x 2 (p), x»(p)) for each
point p of E . Elementary calculus does not always make a sharp distinc
tion between the numbers p u p 2 , p 3 and the functions x 1} x 2 , x 3 . Indeed the
analogous distinction on the real line may seem pedantic, but for higher
dimensional spaces such as E 3 , its absence leads to serious ambiguities.
(Essentially the same distinction is being made when we denote a function
on E 3 by a single letter/, reserving /(p) for its value at the point p.)
We assume that the reader is familiar with partial differentiation and
its basic properties, in particular the chain rule for differentiation of a
composite function. We shall work mostly with firstorder partial deriva
tives df/dx, df/dy, df/dz and secondorder partial derivatives d 2 f/dx 2 ,
d f/dx dy, • ■ • . In a few situations, third, and even fourth, order deriva
tives may occur, but to avoid worrying about exactly how many derivatives
we can take in any given context, we establish the following definition.
1.3 Definition A realvalued function / on E 3 is differentiable (or, in
finitely differentiable, or of class C°°) provided all partial derivatives of
/, of all orders, exist and are continuous.
Differentiable realvalued functions / and g may be added and multi
plied in a familiar way to yield functions that are again differentiable and
realvalued. We simply add and multiply their values at each point— the
formulas read
(/+<7)(p) =/(p) + <7(p), (fg)(p) =/(pMp).
The phrase "differentiable realvalued function" is unpleasantly long.
Hence we make the convention that unless the context indicates otherwise,
"function" shall mean "realvalued function," and (unless the issue is
explicitly raised) the functions we deal with will be assumed to be dif
Sec. 1] EUCLIDEAN SPACE 5
ferentiable. We do not intend to overwork this convention; for the sake
of emphasis the words "differentiable" and "realvalued" will still appear
fairly frequently.
Differentiation is always a local operation : To compute the value of the
function df/dx at a point p of E 3 , it is sufficient to know the values of / at
all points q of E 3 which are sufficiently near p. Thus Definition 1.3 is
unduely restrictive; the domain of /need not be the whole of E , but need
only be an open set of E 3 . By an open set of E we mean a subset of E
such that if a point p is in 0, then so is every other point of E that is
sufficiently near p. (A more precise definition is given in Chap. II.) For
example, the set of all points p = (pi, pi, pz) in E 3 such that p x > is an
open set, and the function yz log x defined on this set is certainly differ
entiable, even though its domain is not the whole of E . Generally speaking,
the results in this chapter remain valid if E 3 is replaced by an arbitrary
open set of E .
We are dealing with threedimensional Euclidean space for no better
reason than that this is the dimension we use most often in later work. It
would be just as easy to work with Euclidean nspace E", for which the
points are ntuples p = (pi, • • • , p n ), and which has n natural coordinate
functions x x , • • • , x n . All the results in this chapter are valid for Euclidean
spaces of arbitrary dimensions, although we shall rarely take advantage
of this except in the case of the Euclidean plane E 2 . In particular, the re
sults are valid for the real line E 1 = R. Many of the concepts introduced
are specifically designed to deal with higher dimensions, however, and are
thus apt to be overelaborate when reduced to dimension 1.
EXERCISES
1. Let / = x 2 y and g = y sin z be functions on E 3 . Express the following
functions in terms of x, y, z:
<•>"■• w %< + %'•
b_
by dz N ~' by
(C)?M. (d)i(sin/)
2. Find the value of the function / = x 2 y — y 2 z at each point :
(a) (1,1,1). ' (b) (3, 1,).
(c) (a, 1,1 a). (d) (t,t 2 ,f).
6 CALCULUS ON EUCLIDEAN SPACE [Chap. I
3. Express df/dx in terms of x, y, and z if
(a) / = x sin (xy) + y cos (xz).
(b) f = sing, g = e\ ft = x 2 + ?/ 2 + z 2 .
4. If 0i, 02, ^3 and h are realvalued functions on E 3 , then
/ = h(gi,g 2 ,g 3 )
is the function such that
/(p) = A(0i(p)t 02 (p), 03 (p)) for all p.f
Express d//dx in terms of x, y, and z, if h = x 2 — yz and
(a) / = A(s + y, i/ 2 , x + z).
(b) /=/*(**, e* + *,e*).
(c) / = h(x, —x, x).
2 Tangent Vectors
Intuitively, a vector in E is an oriented line segment, or "arrow." Vectors
are used widely in physics and engineering to describe forces, velocities,
angular momenta, and many other concepts. To obtain a definition that
is both practical and precise, we shall describe an "arrow" in E 3 by giving
its starting point p and the change, or vector v, necessary to reach its end
point p + v. Strictly speaking, v is just a point of E 3 .
2.1 Definition  A tangent vector \ p to E consists of two points of E 3 :
its vector part v and its point of application p.
We shall always picture \ p as the arrow from the point p to the point
p + v. For example, if p = (1,1,3) and v = (2,3,2), then \ p runs
from (1,1,3) to (3,4,5) as in Fig. 1.1.
We emphasize that tangent vectors are equal, \ p = w g , if and only if
they have the same vector part, v == w, and the same point of application,
p = q. Tangent vectors v,, and v g with the same vector part, but differ
ent points of application, are said to be parallel (Fig. 1.2). It is essential
to recognize that \ p and \ q are different tangent vectors if p ^ q. In physics
the concept of moment of a force shows this clearly enough: The same
force v applied at different points p and q of a rigid body can produce
quite different rotational effects.
f A consequence is the identity / = f(x, y, z).
% The term "tangent" in this definition will acquire a more direct geometric mean
ing in Chapter IV.
Sec. 2]
TANGENT VECTORS
E 3
p + v = (3, 4, 5)
P= (1,1,3)
FIG. 1.1
2.2 Definition Let p be a point of E 3 . The set ^(E 3 ) consisting of all
tangent vectors that have p as point of application is called the tangent
space of E 3 at p (Fig. 1.3).
We emphasize that E 3 has a different tangent space at each and every
one of its points.
Since all the tangent vectors in a given tangent space have the same
point of application, we can borrow the vector addition and scalar multi
plication of E 3 to turn T P (E Z ) into a vector space. Explicitly, we define
Vp _j_ Wp to be (v + w) p , and if c is a number we define c(v p ) to be (cv)„.
This is just the usual "parallelogram law" for addition of vectors, and scalar
multiplication by c merely stretches a tangent vector by factor c— reversing
its direction if c < (Fig. 1.4).
These operations on each tangent space T P (E 3 ) make it a vector space
isomorphic to E 3 itself. Indeed it follows immediately from the definitions
above that, for a fixed point p, the function v » \ p is a linear isomorphism
FIG. 1.3
CALCULUS ON EUCLIDEAN SPACE [Chap. I
(**)«■
FIG. 1.4
from E to ^(E ) — that is, a linear transformation which is onetoone
and onto.
A standard concept in physics and engineering is that of force field. The
gravitational force field of the earth, for example, assigns to each point
of space a force (vector) directed at the center of the earth.
2.3 Definition A vector field V on E 3 is a function that assigns to each
point p of E 3 a tangent vector F(p) to E 3 at p.
Roughly speaking, a vector field is just a big collection of arrows, one
at each point of E 3 .
There is a natural algebra of vector fields. To describe it, we first re
examine the familar notion of addition of realvalued functions / and g.
It is possible to add/ and g because it is possible to add their values at each
point. The same is true of vector fields V and W. At each point p, their
values V(p) and W(p) are in the same vector space— the tangent space
^(E 3 )— hence we can add V(p) and W(p). Consequently we can add V
and W by adding their values at each point. The formula for this addition
is thus the same as for addition of functions,
(V+W)(p) = V(p) + W(p).
This scheme occurs over and over again. We shall call it the pointwise
principle: If a certain operation can be performed on the values of two
functions at each point, then that operation can be extended to the func
tions themselves; simply apply it to their values at each point.
For example, we invoke the pointwise principle to extend the operation
of scalar multiplication (on the tangent spaces of E 3 ). If / is a realvalued
function on E 3 and V is a vector field on E 3 , then fV is defined to be the
Sec. 2]
TANGENT VECTORS
vector field on E 3 such that
(fV)(p) =/( P )F(p) for all p.
Our aim now is to determine in a concrete way just what vector fields
look like. For this purpose we introduce three special vector fields which
will serve as a "basis" for all vector fields.
2.4 Definition Let U h £/ 2 , and U 3 be the vector fields on E 3 such that
CTi(p) = (1,0,0),
tfi(p) = (0,1,0),
U z (p) = (0,0, l) p
for each point p of E 3 (Fig. 1.5). We call U\, Ui, U 3 — collectively — the
natural frame field on E .
Thus Ui (i = 1, 2, 3) is the unit vector field in the positive x» direction.
2.5 Lemma If V is a vector field on E 3 , there are three uniquely deter
mined realvalued functions vi, v iy v 3 on E 3 such that
V = villi + V2U2 + vzUz.
The functions Vi, v 2 , v 3 are called the Euclidean coordinate functions of V.
Proof. By definition, the vector field V assigns to each point p a tangent
vector V(p) at p. Thus the vector part of F(p) depends on p, so we write
it 0>i(p), ^(p), V3(p)). (This defines V\, V2, and vz as realvalued functions
on E 3 . ) Hence
V(p) = (vi(p),v 2 (p),v z (p)) p
= »i(p) (1,0,0), + t*(p) (0,1,0), + vt(p) (0,0,1),
= vi(p)C/i( P ) + y 2 (p)f/ 2 (p) + v 3 (p)Uz(p)
for each point p (Fig. 1.6). By our (pointwise principle) definitions this
means that the vector fields V and X) v <Ui have the same (tangent vector)
value at each point. Hence V = ^ ViUi
This last sentence uses two of our stand
ard conventions: X) y *^» means sum over
i = 1, 2, 3; the Halmos symbol () indi
cates the end of a proof.
The tangentvector identity (ai, a?., a 3 ),
= 23°*^«(p) appearing in this proof will
be used very often.
Computations involving vector fields x i
may always be expressed in terms of their
I
FIG. 1.5
10
CALCULUS ON EUCLIDEAN SPACE
[Chap. I
/ ' v 3 (p)Ui(p) 7\
«*(P)0(p)
FIG. 1.6
Euclidean coordinate functions. For example, addition, and multiplication
by a function, are expressed in terms of coordinates by
^ViUi + ^WiUi = £ (Vi + Wi)Ui
Since this is differential calculus, we shall naturally require that the various
objects we deal with be differentiate. A vector field V is differentiable
provided its Euclidean coordinate functions are differentiable (in the sense
of Definition 1.3). From now on, we shall understand "vector field" to
mean "differentiable vector field."
EXERCISES
1. Let v = (2, 1, 1) and w = (0, 1, 3).
(a) At an arbitrary point p, express the tangent vector 3v p  2w p as
a linear combination of Ui(p), *7 2 (p), Uz(p).
(b) For p = (1, 1, 0), make an accurate sketch showing the four
tangent vectors v p , w p , — 2v p , and v p + w p .
2. Let V = xU x + yU 2 and W = 2x 2 U 2  ZJ%. Compute the vector field
W — xV, and find its value at the point p = ( — 1, 0, 2).
3. In each case, express the given vector field V in the standard form
£ Villi.
(a) 2z 2 £/ 1 = 7V + xyU*.
(b) V(p) = (p lt p 3  p u 0) p for all p.
(c) V = 2(xUi + yUt)  x{U x  y 2 U 3 ).
Sec. 3] DIRECTIONAL DERIVATIVES 11
(d) At each point p, 7(p) is the vector from the point (pi, p 2 , pz) to
the point (1 + p h p 2 p 3 , p 2 ).
(e) At each point p, 7(p) is the vector from p to the origin.
4. If V = y Ui — x Uz and W = x 2 U\ — zUz, find functions / and g such
that the vector field fV + gW can be expressed in terms of £/ 2 and Uz
only.
5. Let 7x = C/i  xt/ 3 , 7 2 = £/ 2 , and 7 3 = xt/i + C/ 3 .
(a) Prove that the vectors Fi(p), F 2 (p), F 3 (p) are linearly inde
pendent at each point of E 3 .
(b) Express the vector field xU\ + yVi + zllz as a linear combination
of 7i, 7 2 , 7,.
3 Directional Derivatives
Associated with each tangent vector \ p to E is the straight line t — > p + iv
(see Example 4.2). If / is a differentiable function on E 3 , then t — >/(p + t\)
is an ordinary differentiable function on the real line. Evidently the deriva
tive of this function at t = tells the initial rate of change of / as p moves
in the v direction.
3.1 Definition Let / be a differentiable real valued function on E , and
let v p be a tangent vector to E . Then the number
y P [f] =(/(p + «v))«
is called the derivative of f with respect to v p .
This definition appears in elementary calculus with the additional
restriction that v p be a unit vector. Even though we do not impose this
restriction, we shall nevertheless refer to v p f/] as a directional derivative.
For example, we compute v p f/]for the function/ = x 2 yz, with p = (1, 1, 0)
and v = (1,0, 3). Then
p + tv = (1, 1,0) + *(1,0, 3) = (1 + t, 1, 30
describes the line through p in the v direction. Evaluating / along this
line, we get
/(p + tv) = (1 + 2 l (3*) = St  Qt 2  St\
Now
(Ap + «v)) = 3 12t9t 2 ;
■'(./
12 CALCULUS ON EUCLIDEAN SPACE [Chap. I
hence at t = 0, we find \ p [f] = — 3. Thus, in particular, the function / is
(initially) decreasing as p moves in the v direction.
The following lemma shows how to compute \ p \f] in general, in terms
of the partial derivatives of / at the point p.
3.2 Lemma If y p = (vi, v 2 , v*) p is a tangent vector to E 3 , then
▼»[/] = £ v t £ ( P ).
Proof. Let p = (pi,p2,p3) ; then
p + t\ = (pi + tv h p 2 + tv*, p 3 + tv z ).
We use the chain rule to compute the derivative at t = of the function
/(p + tv) = /(pi + tvi, p 2 + tv 2 , p 3 + to*).
Since
d
,. (p% + to.) = v»,
we obtain
v p [/] = j t (/(p + *v)) « = £ ^ (p)».. I
Using this lemma, we recompute v p [/] for the example above. Since
/ = x 2 yz, we have
Thus at the point p = (1,1,0),
( P ) = 0> (p)=0, and (p) = l.
Then by the lemma,
v,[/]= + 0+ (3)1 = 3,
as before.
The main properties of this notion of derivative are given in Theorem 3.3.
3.3 Theorem Let / and g be functions on E 3 , v p and w p tangent vectors,
a and b numbers. Then
(1) (av p + bw,)[/l = av p [/] + bw p [f\.
(2) v p [af + bg] = av p [/] + ftvpfo].
(3) v,\fg] = y P Lf]^(p) +/(p)v,fe].
Sec. 3] DIRECTIONAL DERIVATIVES 13
Proof. All three properties may be deduced easily from the preceding
lemma. For example, we prove (3). By the lemma, if v = {v\,vi,vz), then
v,[/<7l = & ^ (P).
But
Hence
dxi dXi dXi
(Z*£(P>><P>+/<P>(2>.*;<r>)
= v,l/l«f(p) +/(p) •▼,!»]. I
The first two properties in the preceding theorem may be summarized
by saying that \ p [f\ is linear in v p and in /. The third property, as its proof
makes clear, is essentially just the usual Leibniz rule for differentiation of
a product. No matter what form differentiation may take, it will always have
suitable linear and Leibnizian properties.
We now use the pointwise principal to define the operation of a vector
field V on a function f. The result is the realvalued function V[f] whose
value at each point p is the number V(p)\f\ — that is, the derivative of/
with respect to the tangent vector F(p) at p. This process should be no
surprise, since for a function / on the real line, one begins by defining the
derivative of fata point — then the derivative function df/dx is the function
whose value at each point is the derivative at that point. Evidently the
definition of V\f] is strictly analogous to this familiar process. In particular,
if Ui, U%, U% is the standard frame field on E 3 , then Ui[f] = df/dxi. This
is an immediate consequence of Lemma 3.2. For example, ZJ\ (p) = (l,0,0) p ;
hence
Ui(p)\f\ =  (/(pi + t, p*, p 3 ))«=o,
which is precisely the definition of (df/dxi) (p). This is true for all points
P = (Pi,P2,Ps); hence Uitf] = df/dxi.
We shall use this notion of directional derivative more in the case of
vector fields than for individual tangent vectors.
3.4 Corollary If V and W are vector fields on E , and /, g, h are real
valued functions, then
14 CALCULUS ON EUCLIDEAN SPACE [Chop. I
(1) (fV + gW)[h] = fV[h] + gW[h].
(2) V[af + bg] = aV[f] + bV\g], for all real numbers a and b.
(3) V\fg]= V[f\g + fV\g].
Proof. The pointwise principle guarantees that to derive these properties
from Theorem 3.3 we need only be careful about the placement of paren
theses. For example, we prove the third formula. By definition, the value
of the function V[fg] at p is V(p)\fg]. But by Theorem 3.3 this is
V(p)\f\g(p) +/(p)7(p)fo] = F[/](p).<7(p) +/(p)7M(p)
= (V\f\'9+fV\g])(p). I
If the use of parentheses here seems extravagant, we remind the reader
that a meticulous proof of Leibniz' formula
— (f ) = ^l» \ f.^
dx dx dx
must consist of exactly the same shifting of parentheses.
Note that the linearity of V[f] in V and / is for functions as "scalars"
in the first formula in Corollary 3.4 but only for numbers as "scalars" in
the second. This stems from the fact that/F signifies merely multiplication,
but V\f] is differentiation.
The identity Ui[f] = df/dxi makes it a simple matter to carry out explicit
computations. For example, if V = xUi — y 2 U s and/ = x 2 y + z s , then
V[f] = xUAx'y] + xUA* 3 }  y 2 U s [x 2 y]  y 2 U 3 [z 3 }
= x(2xy) +   y 2 (Sz 2 ) = 2x 2 y  3y 2 z 2 .
3.5 Remark Since the subscript notation \ p for a tangent vector is
somewhat cumbersome, from now on we shall frequently omit the point
of application p from the notation. This can cause no confusion, since v
and w will always denote tangent vectors, and p and q points of E 3 . In
many situations (for example, Definition 3.1) the point of application is
crucial, and will be indicated by using either the old notation v,, or the
phrase "a tangent vector v to E 3 at p."
EXERCISES
1. Let \ p be the tangent vector to E 3 for which v = (2, —1, 3) and
p = (2, 0, —1). Working directly from the definition, compute the
directional derivative v p [/], where
(a) / = yz. (b) / = x\ (c) / = e cos y.
Sec. 4] CURVES IN E 3 15
2. Compute the derivatives in Ex. 1 using Lemma 3.2.
3. Let V = y 2 Ux  xU z , and let/ = xy,g = 2 3 . Compute the functions
(a) 7[fl. (c) 7[M (e) F[f + </ 2 ].
(b) V\g\. (d) /7fe]  flf7[/]. (f) V[V\J]l
4. Prove the identity 7=2 Vfo]l7i, where xi, a*, x 3 are the natural
coordinate functions. (Hint: Evaluate V = 2 ViU t on xy.)
5. If 7[fl = WLfl for every function / on E 3 , prove that V = W.
4 Curves in E 3
Let I be an open interval in the real line R. We shall interpret this liberally
to include not only the usual type of open interval a < t < b (a, b real
numbers), but also the types a < t (a half line to +«),*< 6 (a half line
to — oo ), and also the whole real line.
One can picture a curve in E 3 as a trip taken by a moving point a. At
each "time" t in some open interval, a is located at the point
a(t) = (ai(0,« 2 (0, «s(0)
in E 3 . In rigorous terms then, a is a function from / to E 3 , and the real
valued functions a u a 2 , a 3 are its Euclidean coordinate junctions. Thus we
write a = (ai,a2,a8), meaning, of course, that
a(jt) = (ai(0,« 2 (0,«3(0)
for all •* in the interval I. We define the function a to be differentiate
provided its (realvalued) coordinate functions are differentiate in the
usual sense.
4.1 Definition A curve in E 3 is a differentiate function a: I —*■ E from
an open interval I into E .
We shall give several examples of curves, which will be used in Chapter
II to experiment with results on the geometry of curves.
4.2 Example
(1) Straight line. A line is the simplest type of curve in Euclidean space;
its coordinate functions are linear (in the sense t — > at + 6, not in the
homogeneous sense t ^ at). Explicitly, the curve a: R * E 3 , such that
a (t) = p + tq = (pi + tqi, p 2 + tq 2 , p 3 + tqt) (q ^ 0)
is the straight line through the point p = a(0) in the q direction.
(2) Helix. (Fig. 1.7). The curve t > (a cos t, a sin t, 0) travels around
w
CALCULUS ON EUCLIDEAN SPACE
[Chap. I
FIG. 1.7
FIG. 1.8
a circle of radius a > in the xy plane of E 3 . If we allow this curve to rise
(or fall) at a constant rate, we obtain a helix a: R
formula
E , given by the
a {t) = (a cos t, a sin t, bt)
where a > 0, 6 ^ 0. (We shall always use the term helix to mean right
circular helix. )
(3) Let
<x(t) = (2 cos 2 /, sin 2t, 2 sin t) for < t < w/2.
This curve a has a noteworthy property: Let C be the cylinder in E 3 con
structed on the circle in the xy plane with center at (1,0,0) and radius 1.
Then a follows the route sliced from C by the sphere S with radius 2 and
center at the origin (Fig. 1.8).
(4) The curve a: R — > E 3 such that
a(t) = (e\e~\y/2t)
shares with the helix in (2) the property of rising constantly. However, it
lies over the hyperbola xy = 1 in the xy plane instead of a circle.
(5) The curve a: R ► E 3 such that
a(t) = (St  t\St 2 ,3t + t s ).
If the coordinate functions of a curve are simple enough, the shape the
curve has in E 3 can be found, at least approximately, by the bruteforce
procedure of plotting points. We could get a reasonable picture of this curve
for ^ t ^ 1 by computing a(t) for t = 0, Ho, H, Ho, 1.
If we visualize a curve a in E 3 as a moving point, then at every time t
Sec. 4] CURVES IN E 3 17
there is a tangent vector at the point a(t) which gives the instantaneous
velocity of a at that time.
4.3 Definition Let a: / — ► E be a curve in E with a = (a 1} an, 013). For
each number t in /, the velocity vector of a at t is the tangent vector
«'(0 (£<o,^<0.^(«)
dt dt dt }a{t)
at the point a(t) in E a (Fig. 1.9).
We interpret this definition geometrically as follows. The derivative
at t of a realvalued function / on R is given by
dt Ato At
This formula still makes sense if/ is replaced by a curve a = (an, 0:2,0:3).
In fact,
<«(« + M)  «(()) = • " (t + M) ~ aM
c
At" V At
a 2 « + At)  a 2 (t) a t (t + At)  a»(t) \
At ' At )'
This is the vector from a(t) to a(t + At), scalar multiplied by l/At (Fig*
1.10).
Now as At gets smaller, a(t " Af) approaches a(0> and in the limit as
A£ — > 0, we get a vector tangent to the curve a at the point a(t), namely,
(dai/dt(t), da 2 /dt(t), da 3 /dt(t)). As the figure suggests, the point of
application of this vector must be the point a(t). Thus the standard limit
operation for derivatives gives rise to our definition of the velocity of a
curve.
<*'(*)
FIG. 1.9
18
CALCULUS ON EUCLIDEAN SPACE
1
[Chap. I
(a(t + At)  a(t))
FIG. 1.10
An application of the identity
to the velocity vector a (t) at t yields the alternative formula
«'(*) = E^(0^(«(0).
For example, the velocity of the straight line a(t) = p + tq is
«'(0 = (qi,qt,qi)aw= q«<«).
The fact that a is straight is reflected in the fact that all its velocity vectors
are parallel; only the point of application changes as t changes.
For the helix
the velocity is
a(t) = (a cos t, a sin t, bt),
a (0 = (—a sin t, a cos t, b)
a(t)
The fact that the helix rises constantly is shown by the constancy of the
z coordinate of a (t).
Given a curve a, one can construct many new curves which follow the
same route as a, but travel at different speeds.
4.4 Definition Let / and J be open intervals in the real line R. Let a:
I + E be a curve and let h: J ► / be a differentiate (realvalued) func
tion. Then the composite function
= a(h): J+E 3
is a curve called the reparametrization of a by h.
At each time s in the interval J, the curve p is at the point (*) = « (A (a) )
reached by the curve a at time A(s) in the interval / (Fig. 1.11). Thus
does follow the route of a, but fi generally reaches a given point on the route
at a different time than a does. In practice, to compute the coordinates of
0, one simply substitutes t = h(s) in the coordinates ai(t), a 2 (t), a »(t) of
Sec. 4] CURVES IN E 3 19
FIG. 1.11
a. For example, suppose a(t) = (\/t, ty/l, 1 — t) on J: < t < 4. If
h(s) = s 2 on J: < s < 2, then
0(s) = a(Ms)) = «(s 2 ) = (s, s\ 1  s 2 ).
Thus the curve a: I > E 3 has been reparametrized by to to yield the curve
0: J>E 3 .
The following lemma relates the velocities of a curve and of a repara
metrization.
4.5 Lemma If /3 is the reparametrization of a by h, then
/?'(«) = (dh/ds)(s)a(h(s)).
Proof. If a = (ai, a2, 03), then
j8(s) = a(ft(*)) = (ai(A(s)),a,(A(*)),a,(A(«)).
Using the "prime" notation for derivatives, the chain rule for a composi
tion of realvalued functions / and g reads (g(J)Y = g'(J)'f Thus in tne
case at hand, we obtain
aM'is) = ai '(h(s))h'(s).
By the definition of velocity, this yields
ft {*) = «(*)'(«)
= (a/ (*(«)) •*'(«),«* (*(«))•* («),«» (*(•))•* («))
= *'(*)«'(*(«))■ I
According to this lemma, to obtain the velocity of a reparametrization
of a by h, first reparametrize a by h, then scalar multiply by the derivative
of ft.
Since velocities are tangent vectors, we can take the derivative of a
function with respect to a velocity.
4.6 Lemma Let a be a curve in E 3 and let / be a differentiable function
on E 3 . Then
«'(0W  d ^ }1 (0.
20
CALCULUS ON EUCLIDEAN SPACE
[Chap. I
Proof. Since
we conclude from Lemma 3.2 that
«'(Wi = Z^(«(0)^«).
dXi
dt
But the composite function /(a) may be written /(ai,a 2 ,a 3 ), and the
chain rule then gives exactly the same result for the derivative of f(a). 
By definition, a' (Of/] is the rate of change of / along the line through
a(t) in the a (t) direction (Fig. 1.12). (If a (t) * 0, this is the tangent
line to a at a (t) ; see Exercise 9.) The lemma shows that this rate of change
is the same as that of / along the curve a itself.
Since a curve a: I — ► E 3 is a function, it makes sense to say that a is one
toone; that is, a(t) = a(ti) only if t — h. Another special property of
curves is periodicity: A curve a: R > E 3 is periodic if there is a number
p > such that a(t + p) = a(t) for all t— and the smallest such number
p is then called the period of a.
From the viewpoint of calculus, the most important condition on a
curve a is that it be regular, that is, have all velocity vectors different from
zero. Such a curve can have no corners or cusps.
The following remarks about curves (offered without proof) are not an
essential part of our exposition, but will be of use in Chapter IV. We con
sider, in the case of the plane E 2 , another familiar way to formulate the
concept of "curve." If / is a differentiable realvalued function on E 2 , let
C:f=a
be the set of all points p in E 2 such that /(p) = a. Now if the partial
derivatives df/dx and df/dy are never simultaneously zero at any point of
C, then C consists of one or more separate "components" which we shall
^
/.
\\
//
\s
//
\\
/ /
\ s
/ 1
\ \
/ I
^21 \
/ fc.
j •
\ 1
/ /
N \
1 /
\ \
1 /
\ \
/ /
\ \
//
\\
//
>X
V
V
FIG. 1.12
FIG. 1.13
Sec. 4] CURVES IN E 3 21
call Curves^ Thus, for example, C: x + y 2 = r 2 is the circle of radius r
centered at the origin of E 2 , and the hyperbola C: x — y 2 — r 2 splits into
two Curves ("branches") d and C 2 as in Fig. 1.13.
Every Curve C is the route of many regular curves a, called parametriza
tions of C. If C is a closed Curve, then it has a periodic parametrization
a: R — > C. For example, the curve
a(t) = (r cos t, r sin t)
is a wellknown periodic parametrization of the circle given above. If C
is a Curve that is not closed (sometimes called an arc), then every para
metrization £: / — > C is onetoone. For example,
/8(0 = (r cosh t, r sinh £)
parametrizes the branch, x > 0, of the hyperbola given above.
EXERCISES
1. Compute the velocity vector of curve (3) in Example 4.2 for arbitrary
t, and at t = x/4.
2. Plot curve (5) in Example 4.2 by the method there suggested. On
the sketch show the velocity vectors at t = 0, , 1.
3. Find the coordinate functions of the curve = a(h), where a is curve
(3) in Example 4.2 and h is the function on J: < s < 1 such that
h(s) = sin 1 s.
4. Find the (unique) curve a such that a(0) = (1,0, —5) and a (t) =
(t\t,e<).
5. Find a straight line passing through the points (1, —3, —1) and
(6, 2, 1). Does this line meet the line through the points (—1, 1, 0)
and (5, 1, 1)?
6. Deduce from Lemma 4.6 that in the definition of directional derivative
(Definition 3.1) the straight line t — ► p + tv may be replaced by any
curve a with initial velocity v p , that is, such that a(0) = p and a (0) =
v p .
7. (Continuation). Show that the curves given by (/, 1 + t 2 , t), (sin t, cos t,
t), and (sinh t, cosh t, t) all have the same initial velocity \ p .
If / = x 2 — y + z, compute v p \f] by evaluating / on each of the curves.
8. Let h(s) = log s on J: s > 0. Reparametrize curve (4) in Example 4.2
t In this section (only) we use a capital C to distinguish this notion from a curve
a:/^E 3 .
22 CALCULUS ON EUCLIDEAN SPACE [Chap. I
using h. Check the equation in Lemma 4.5 in this case by computing
each side separately.
9. For a fixed t, the tangent line to a regular curve a at a (t) is the straight
line u — ► a(t) + ua'(t), where we delete the point of application of
a (t). Find the tangent line to the helix a(t) = (2 cos t, 2 sin t, t) at
the points <*(0) and a(x/4).
10. Sketch the following Curves in E 2 and find parametrizations for each.
(a) (7:4/ + y 2 = 1. (c) C: y = e x .
(b) C: Zx + 4y = 1. (d) C: z 2/3 + t/ 2/3 = 1, x > 0, j, > 0.
5 1 Forms
If / is a realvalued function on E 3 , then in elementary calculus one defines
the differential of / to be
df=~dx+ d ldy+ d /dz.
dx dy * dz
It is not always made clear exactly what this formal expression means.
In this section we give a rigorous treatment using the notion of 1form,
and these will tend to appear at crucial moments in our later work.
5.1 Definition A 1form <j> on E 3 is a realvalued function on the set of
all tangent vectors to E 3 such that <j> is linear at each point, that is,
4>(a\ + 6w) = cuf>(\) + b<f>(w)
for any numbers a, b and tangent vectors v, w at the same point of E 3 .
We emphasize that for every tangent vector v to E 3 , a 1form <f> defines
a real number <f> (v) ; and for each point p in E 3 , the resulting function
<j> p : T P (E ) — »• R is linear. [Thus at each point p, <f> p is an element of the
dwl space of ^(E 3 ). In this sense the notion of 1form is dual to that of
vector field.]
The sum of 1 forms <t> and \p is defined in the usual pointwise fashion
(<t> + ^) (v) = <£(v) + ^(v) for all tangent vectors v.
Similarly if / is a real valued function on E 3 and </> is a 1form, then fy>
is the 1form such that
(fo)(vp) =/(p)*(v,)
for all tangent vectors v p .
There is also a natural way to evaluate a 1form <j> on a vector field V to
obtain a realvalued function 4>(V): At each point p the value of <f>(V)
Sec. 51 1 FORMS 23
is the number <f>(V(p)). Thus a 1form may also be viewed as a machine
which converts vector fields into realvalued functions. If <f> (V) is differenti
able whenever V is, we say that <£ is differentiable. As with vector fields,
we shall always assume that the 1 forms we deal with are differentiable.
A routine check of definitions shows that <j>(V) is linear in both and V;
that is,
*(fV + gW) = f<t>(V) +g<t>{W)
and
(ft + 0*)(F) = f<t>(V) +g + (V)
where / and g are functions.
Using the notion of directional derivative, we now define a most impor
tant way to convert functions into 1 forms.
5.2 Definition If / is a differentiable realvalued function on E 3 , the
differential df of / is the 1form such that
df(v p ) = y p [f] for all tangent vectors y p .
In fact, df is a, 1form, since by definition it is a realvalued function on
tangent vectors, and by (1) of Theorem 3.3 it is linear at each point p.
Clearly df knows all rates of change of / in all directions on E 3 , so it is not
surprising that differentials are fundamental to the calculus on E 3 .
Our task now is to show that these rather abstract definitions lead to
familiar results when expressed in terms of coordinates.
5.3 Example 1 Forms on E 3 . (1) The differentials dx lf dx 2 , dx 3 of the
natural coordinate functions. Using Lemma 3.2 we find
dxi(\ p ) = v p [ Xi ] = Yj>i jt 1 (p) = Yji s n = v i
j OXj j
where «, 7 is the Kronecker delta (0 if i 9± j, 1 if i = j). Thus the value of dx {
on an arbitrary tangent vector v p is the ith coordinate Vi of its vector part — and
does not depend at all on the point of application p.
(2) The 1form f = fi dxi + / 2 dx 2 + f 3 dx 3 . Since dxi is a 1form, our
definitions show that ^ is also a 1form for any functions f u / 2 , / 3 . The value
of ^ on an arbitrary tangent vector v p is
*(▼») = (£/«**)(▼,) = Hfi(p)dx i (y) = E/<(p)»*
The first of these examples shows that the 1 forms dx u dx 2 , dx 3 are the
analogues for tangent vectors of the natural coordinate functions xi, z 2 , x 3
for points. Alternatively, we can view dx u dx 2 , dx 3 as the "duals" of the
natural unit vector fields U u U 2 , U 3 . In fact, it follows immediately from
(1) above that the function dXi(Uj) has the constant value 5 t7 .
24 CALCULUS ON EUCLIDEAN SPACE [Chap. I
We shall now show that every 1form can be written in the concrete
manner given in (2 ) above.
5.4 Lemma If <£ is a 1form on E , then <f> = 52 fi dxi, where / t = <f>(Ui).
These functions /i, / 2 , / 3 are called the Euclidean coordinate functions of <f>.
Proof. By definition a 1form is a function on tangent vectors; thus <f>
and 52 /*' dxi are equal if and only if they have the same value on every
tangent vector v p = 52 ViUi(p). In (2) of Example 5.3 we saw that
(2/.dx.)(vp) = 52/*(pH
On the other hand,
*(v,) =0(X>;£A(p)) = 5>**»(^(p)) = I><A(p)
since /» = <j>(Ui). Thus <f> and 52 fidxt do have the same value on every
tangent vector. 
This lemma shows that a 1form on E is nothing more than an expres
sion fdx + g dy + h dz, and such expressions are now rigorously defined
as functions on tangent vectors. Let us now show that the definition of
differential of a function (Definition 5.2) agrees with the informal defini
tion given at the start of this section.
5.5 Corollary If / is a differentiate function on E 3 , then
Proof. The value of 52 (df/dxi) dxi on an arbitrary tangent vector
v p is 52 (df/dXi)(p)vi. By Lemma 3.2 df{\ p ) = \ p [f] is the same. Thus
the 1 forms df and 52 W/dx,) dxi are equal. 
Using either this result or the definition of d, it is immediate that
dV+g) = df + dg.
Finally we determine the effect of d on products of functions and on compo
sitions of functions.
5.6 Lemma Let fg be the product of differentiate functions / and g
on E 3 . Then
d(fg) = gdf + fdg.
Proof. Using Corollary 5.5, we obtain
«*>E'4?*<z(£;. + /i>.
Sec. 5] 1FORMS 25
5.7 Lemma Let /: E 3 — > R and fc: R — > R be differentiable functions,
so the composite function h (/) : E 3 — ► R is also differentiable. Then
<*(*(/)) =h'(f)df.
Proof. (The prime here is just the ordinary derivative, so h (/) is again
a composite function, from E 3 to R.) The usual chain rule for a composite
function such as h (/) reads
m» . kvff
dXi dXi
Hence .
d(Hf)) = £ *<^ <&* = £ fc'(/> ^ <fe, = h'(f)df. 
To compute df for a given function / it is almost always simpler to use
these properties of d rather than substitute in the formula of Corollary
5.5. Then from df we immediately get the partial derivatives of / and, in
fact, all its directional derivatives. For example, suppose
/= (x 2  l)y + Q/ 2 + 2)z.
Then by Lemmas 5.6 and 5.7,
df = (2x dx)y + (x 2  1) dy + (2y dy)z + (y 2 + 2) dz
= 2xy dx + (x 2 + 2 yz  \) dy + (y 2 + 2) dz
d//dx d//3y d//da
Now use the rules above to evaluate this expression on a tangent vector
v . The result is
v,[/l = df(y p ) = 2pip 2 fi + (pi 2 + 2p 2 p 3  l)w« + (P2 2 + l)w».
EXERCISES
1. Let v = (1, 2, —3) and p = (0, —2, 1). Evaluate the following
1 forms on the tangent vector v p .
(a) y 2 dx. (b) zdy — y dz. (c) (z 2 — \)dx — dy + z 2 dz.
2. If <t> = 52 /f dx» an d V = 22 v *^«> show that the 1form </> evaluated
on the vector field V is the function 4>(V) = 22 /»' y <
3. Evaluate the 1form <£ = x 2 dx — y 2 dz on the vector fields
V = xU x + ?/t/ 2 + z£/ 3 ,
TF = xy{U,  U t ) + y«(tfi  Ut), and (l/z)F + (l/y)W.
26 CALCULUS ON EUCLIDEAN SPACE [Chap. I
4. Express the following differentials in terms of df:
(a) d(f). (b) d(VJ), where/ > 0. (c) d(log(l +/ 2 )).
5. Express the differentials of the following functions in the standard
form ^fidxi.
(a) {x 2 + y 2 + i) m . (b) tan^O//*).
6. In each case compute the differential of / and find the directional de
rivative v p [/], for v p as in Ex. 1.
(a) / = xy 2 — yz. (b) / = xe yz . (c) / = sin {xy) cos {xz).
7. Which of the following are 1 forms? In each case <f> is the function on
tangent vectors such that the value of <f> on (vi, v 2 , v 3 ) p is
(a) Vi — vs. (c) vtps + v 2 pi. (e) 0.
{b) Pl p z . {d) v p [x 2 + /}. (f) (pi) 2 .
In case <j> is a 1form, express it as 23 /*" ^»
8. Prove Lemma 5.6 directly from the definition of d — without using
Corollary 5.5.
9. A 1form is zero at a point p provided 4>{v P ) = for all tangent
vectors at p. A point at which its differential df is zero is called a
critical point of the function /. Prove that p is a critical point of /
if and only if
d l ( p ) = d l ( p ) = d l ( p ) = o.
dx p/ by p bz vp/
Find all critical points of / = (1 — x 2 )y + (1 — y )z.
{Hint: Find the partial derivatives of / by computing df. )
10. {Continuation) . Prove that the local maxima and local minima of /
are critical points of /. (/ has a local maximum at p if /(q) ^ /(p)
for all q near p.)
11. It is sometimes asserted that df is the linear approximation of A/.
(a) Explain the sense in which {df) (v p ) is the linear approximation
of/(p + v) /(p).
(b) Compute the exact and approximate values of / (0.9, 1.6, 1.2) —
/(l, 1.5, 1), where/ = x 2 y/z.
6 Differential Forms
The 1 forms on E 3 are part of a larger system called the differential forms
on E 3 . We shall not give as rigorous an account of differential forms as we
did of 1 forms, for o"r use of the full system will be limited to Section 8
Sec. 6] DIFFERENTIAL FORMS 27
of Chapter II. Roughly speaking, a differential form on E 3 is an expression
obtained by adding and multiplying realvalued functions and the differ
entials dxi, dx2, dxz of the natural coordinate functions of E . These two
operations obey the usual associative and distributive laws; however, the
multiplication is not commutative. Instead it obeys the
alternation rule: dxidxj = —dxjdxi (1 ^ i, j ^3).
This rule appears — although rather inconspicuously — in elementary
calculus (see Exercise 9).
A consequence of the alternation rule is the fact that "repeats are zero,"
that is, dxi dxi = 0, since if i — j the alternation rule reads
If each summand of a differential form contains pdxi's (p = 0, 1, 2, 3),
the form is called a pform, and is said to have degree p. Thus, shifting to
dx,dy,dz, we find
A 0form is just a differentiable function /.
A 1form is an expression fdx f g dy + hdz, just as in the preceding
section.
A 2 form is an expression / dx dy + g dx dz + h dy dz.
A 3 form is an expression / dx dy dz.
We already know how to add 1 forms: simply add corresponding coef
ficient functions. Thus, in index notation,
2 fi dxi f X 9i dxi = 53 (fi + ffi) dxi.
The corresponding rule holds for 2forms or 3forms.
On threedimensional Euclidean space, all pforms with p > 3 are zero.
This is a consequence of the alternation rule, for a product of more than
three dxi's must contain some dx t twice, but repeats are zero, as noted
above. For example, dxdydxdz = —dxdxdydz = 0, since dxdx = 0.
As a reminder that the alternation rule is to be used, we denote this multi
plication of forms by a wedge a . (However, we do not bother with the
wedge when only products of dx, dy, dz are involved.)
6.1 Example Computation of wedge products. (1) Let
tf> = xdx — y dy and ty = z dx + x dz.
Then
4> a ^ = (x dx — y dy) a (z dx + x dz)
= xz dx dx + x 2 dx dz — yz dy dx — yx dy dz.
28 CALCULUS ON EUCLIDEAN SPACE [Chap. I
But
dx dx = 0, and dy dx = — dx dy.
Thus
4> a \f/ = yz dx dy + x 2 dx dz — xy dy dz.
In general, the product of two 1 forms is a 2form.
(2) Let <f> and \f/ be the 1 forms given above and let
9 = z dy.
Then
a $ a 4/ = yz" dy dx dy + x 2 z dy dx dz — xyz dy dy dz.
Since dy dx dy and dy dy dz each contain repeats, both are zero. Thus
a a \f/ = —x 2 z dx dy dz.
(3 ) Let <f> be as above, and let 77 be the 2form y dx dz + x dy dz. Omitting
forms containing repeats, we find
$ a 77 = x 2 dxdy dz — y 2 dy dx dz = (x 2 + y 2 ) dx dy dz.
It should be clear from these examples that the wedge product of a
pform and a qiovm. is a (p + q)iorm. Thus such a product is automatically
zero whenever p + q > 3.
6.2 Lemma If $ and ^ are 1 forms, then
<f> A ^ = — ^ A 0.
Proof. Write
<t> = X /i <&••> ^ = £ 0i fe
Then by the alternation rule,
<t> A 4> — ]£/#/ cfoi dx, =  J] flfj/i cte; dxi = —4* a <f>. 
In the language of differential forms, the operator d of Definition 5.2
converts a 0form / into a 1form df. It is easy to generalize to an operator
(also denoted by d) which converts a p form 77 into a (p + l)form ^77:
One simply applies d (of Definition 5.2) to the coefficient functions of 77.
For example, here is the case p = 1 .
6.3 Definition If <f> = ]£ f { dxi is a 1form on E 3 , the exterior derivative
of <f> is the 2form d4> = ^ dfc a dx { .
If we expand the preceding definition using Corollary 5.5, we obtain the
Sec. 6] DIFFERENTIAL FORMS 29
following interesting formula for the exterior derivative of
<f> = /i dxi + / 2 dxi + fz dx z :
dxi dx%l
\dxi 8x2/ \dxi dx s / \
There is no need to memorize this formula; it is more reliable to simply
apply the definition in each case. For example, suppose
4> = xy dx + x dz.
Then
d4> = d(xy) a dx + d(x ) dz
— (y dx + x dy) a dx + 2x dx dz
= —x dx dy + 2x dx dz.
It is easy to check that the general exterior derivative enjoys the same
linearity property as the particular case in Definition 5.2; that is,
d{a4 + 6^) = a d<t> + b d$,
where </> and \f/ are arbitrary forms and a and 6 are numbers.
The exterior derivative and the wedge product work together nicely.
6.4 Theorem Let / and g be functions, <j> and \f/ 1 forms. Then
(1) d(fg) =dfg+fdg.
(2) d(f<f>) = df a 4>+fd<t>.
(3) d(4> a yp) = d<f> a yp — $ a d\p. f
Proof. The first formula is just Lemma 5.6. We include it to show the
family resemblance of all three formulas. The proof of the second formula
is a simpler variant of that of the third, so we prove only the latter.
Cose 7 . <f> = f dx, *p = g dx. Since
4> a \p = fg dx dx = 0,
we must show that the right side of the equation is also zero. Now
dtf> = df a dx = ^ dy dx + ^ dz dx:
dy dz
hence each term of cUp a ^ has a repeated dx. Thus d$ a \f/ = 0, and simi
larly <£ a dip = 0.
Case 2. <f> = f dx, \p = g dy. Using the formula for d^ computed above,
we get
t As usual, multiplication takes precedence over addition or subtraction, so this
expression should be read as (dtj> A 4) — (<f> A <bf).
30 CALCULUS ON EUCLIDEAN SPACE [Chap. I
dtf> a yf, = iJ dy dx + j dz dx) a g dy
Similarly,
Thus
But
so we get
= + ^ g dz dx dy = g ^ dx dy dz.
dz dz
<f> a dip = fdx a ( ? dx dy + ? dz dy )
\dx dz /
= f f dx dz dy = —f^dxdy dz.
dz dz
* t  <t> a df = \g£ + ^ /J dx dy dz.
4 a $ = fg dx dy,
d(<f> a *) = <*(#) da: dy = ^1 dz dx dy
dz
Hence the formula is proved in this case.
Case 3. The general case. From cases 1 and 2 we know that the formula
is true whenever <f> and ^ are "simple," that is, of the form / du, where u
is x, y, or z. Since every 1form is a sum of simple 1 forms, the general case
follows from the linearity of d and the distributive law for the wedge
product. 
One way to remember the minus sign which occurs in formula (3) of
Theorem 6.4 is to pretend that d is a 1form. To reach ^, d must change
places with <f>; hence the minus sign is consistent with Lemma 6.2.
Differential forms, and the associated concepts of wedge product and
exterior derivative, provide a means of expressing rather complicated rela
tionships in a simple, methodical way. For example, as its proof shows,
the tidy formula
d(<f> a \f/) = d<f> a ^ — <f> a d\{/
involves some rather tricky relations among partial derivatives. Before
Sec. 61 DIFFERENTIAL FORMS 31
forms were invented, it was necessary to struggle through these relations
in many a separate problem, but now we simply apply the general formula.
A variety of interesting applications is given in Flanders [1]. Later on
we shall use differential forms to express the fundamental equations of
geometry.
EXERCISES
1. Let <f> = yz dx + dz, \f/ = sin z dx + cos z dy, £ = dy + z dz. Find the
standard expressions (in terms of dx dy, • • • ,) for
(a) a f, $ a £, % a <t>. (b) dxf>, dxf,, d$.
2. Let <f> = dx/y and ^ = z dy. Check the Leibnizian formula (3) of The
orem 6.4 in this case by computing each term separately.
3. For any function / show that d (df) = 0. Deduce that d{f dg) = df a dg.
A. Simplify the following forms :
(z)d(fdg + gdf). (c) d(fdg a g df).
(b) d{ (/  g)(df + dg)}. (d) d(gfdf) + d(fdg).
5. For any three 1 forms </>; = ^jfijdxj (1 ^ i ^ 3), prove
01 A <f>2 A <f>3 =
/ll /l2 /l3
/21 /22 /23 etei cte 2 dx 3 .
/31 /32 J33
6. If r, #, and z are the cylindrical coordinate functions on E 3 , then
x = r cos &, y = r sin #, z = z. Compute the volume element dx dy dz
of E in cylindrical coordinates. (That is, express dx dy dz in terms of
the functions r, tp, z and their differentials. )
7. For a 2form
77 = f dx dy \ g dx dz \ h dy dz,
the exterior derivative dr\ is defined to be the 3form obtained by replac
ing /, g, and h by their differentials. Prove that for any 1form
<f>,d(d<t>) = 0.
Exercises 3 and 7 show that d 2 = 0, that is, for any form £, d(di) = 0.
(If £ is a 2form, then d(d£) = 0, since its degree exceeds 3.)
8. Classical vector analysis avoids the use of differential forms on E by
converting 1 forms and 2forms into vector fields by means of the follow
ing onetoone correspondences:
£ fi dxi ^ X f&i ^ U dxi dxi — / 2 dxi dx 3 + f x dx 2 dx 3 .
32
CALCULUS ON EUCLIDEAN SPACE
[Chap. I
Vector analysis uses three basic operations based on partial differentia
tion:
Gradient of a function /:
grad/= £^t/;.
dXi
Curl of a vector field V = J^ftUi:
curl V = (°*  d A ) Ul + (*  *) U, + (*  * W
\dx 2 dxj ^ \dx 3 dxj T \d:n dz 2 / 3
Divergence of a vector field V — ^ /»#*:
div 7 = V 3£.
Prove that all three operations may be expressed by exterior deriva
tives as follows:
(a)d/JiLgrad/.
(b) If <f> J1L V, then d0 J^L curl F.
(c) If t, J^L V, then dr, = (div 7) dx dy dz.
9. Let / and g be realvalued functions on E 2 . Prove that
df a dg =
5/
5/
d£
ay
d<7
d<7
3a;
dy
dx cfy.
This formula appears in elementary calculus; show that it implies the
alternation rule.
7 Mappings
In this section we discuss functions from E" to E m . If n = 3 and m = 1,
then such a function is just a realvalued function on E 3 . If n = 1 and
m = 3, then such a function is a curve in E 3 . Although our results will
necessarily be stated for arbitrary m and n, we are primarily interested
only in the three cases:
E 2 ^E 2
E 2
E*
E\
The fundamental observation about a function F: E" — > E m is that it
can be completely described by m real valued functions onE". (We saw
this already in Section 4 for n = \,m = 3.)
Sec. 7] MAPPINGS 33
7.1 Definition Given a function F : E n — > E m , let /i,/ 2 , • • • , /» denote the
realvalued functions on E n such that
F(p) = (/i(p),/ 2 (p), ■■• ,/.(p))
for all points p in E". These functions are called the Euclidean coordinate
junctions of F, and we write F = (fi,fe, • • • , f m )
The function F is differentiable provided its coordinate functions are
differentiable in the usual sense. A differentiable function F: E n — > E TO
is called a mapping from E n to E m .
Note that the coordinate functions of F are the composite functions
fi = Xi(F), where xi, • • • , x m are the coordinate functions of E m .
Mappings may be described in many different ways. For example,
suppose F: E —> E 3 is the mapping F = (x , yz, xy) . Thus
F(p) = (x(p)\ y(p)z{p),x(p)y(p)) for all p.
Now p = (pi, p2, P3), and, by definition of the coordinate functions,
s(p) = pi> y(p) = P2» z (p) = P«
Hence we obtain the following pointwise formula for F:
F (pi, P2, Pz) = {pi, PiPz, P1P2) for all p u p 2 , p 3 .
In particular,
F(l,2,0) = (1,0,2), F(3,l,3) = (9,3,3),
and so on.
In principal, one could deduce the theory of curves from the general
theory of mappings. But curves are reasonably simple, while a mapping,
even in the case E 2 — > E 2 , can be quite complicated. Hence we reverse this
process and use curves, at every stage, to gain an understanding of map
pings.
7.2 Definition If a : I — > E" is a curve in E n and F : E n — > E m is a mapping,
then the composite function = F(a) : I —> E m is a curve in E m called
the image of a under F (Fig. 1.14).
FIG. 1.14
34
CALCULUS ON EUCLIDEAN SPACE
[Chap. I
7.3 Example Mappings. (1) Consider the mapping F: E 3 — > E 3 such
that
F.= (x — y, x + y, 2z).
In pointwise terms then,
F{VuV*>V*) = (pi  P2, Pi + p 2 , 2p 3 ) for all p u p 2 , p % .
Only when a mapping is quite simple can one hope to get a good idea of
its behavior by merely computing its values on some finite number of
points. But this function is quite simple — it is a linear transformation
from E to E . Thus by a wellknown theorem of linear algebra, F is
completely determined by its values on three (linearly independent)
points, say the unit points
ui= (1,0,0) u 2 = (0,1,0) u,= (0,0,1).
(2) The mapping F: E 2 » E 2 such that F(u,v) = (u 2  v 2 , 2uv). (Here
u and v are the coordinate functions of E 2 .) To analyze this mapping, we
examine its effect on the curve a(t) = (r cos t, r sin t), where ^ t ^ 2t.
This curve takes one counterclockwise trip around the circle of radius r
(center at the origin.) The image curve is
0(0 = F(a(t)) = F{r cos t, r sin t) = (r 2 cos 2 t  r 2 sin 2 £, 2r 2 cos t sin t),
with ^ t ^ 2x. Using the trigonometric identities
cos 2t = cos 2 t — sin 2 i, sin 2t = 2 sin t cos t,
we find for = F(a) the formula
0(0 = (r 2 cos2£, r 2 sin 20,
with ^ f ^ 27r. This curve takes to#o counterclockwise trips around the
circle of radius r 2 (center at origin) (Fig. 1.15).
Thus the effect of F is to wrap the plane E 2 smoothly around itself twice —
leaving the origin fixed, since ^(0,0) = (0,0). In this process, each circle
of radius r is wrapped twice around the circle of radius r 2 .
FIG. 1.15
Sec. 7]
MAPPINGS
35
Each time we have defined a new object in this chapter we have pro
ceeded to define a suitable notion of derivative of that object. For example,
the "derivative" of a curve a is its velocity a . Using the notion of velocity
of a curve, we shall now define the derivative F* of a mapping F:
E n — >• E m . F* is going to be a function that assigns to each tangent vector
v to E n at p a tangent vector F*(v) to E m at F(p). We get F*(v) by the
following process: The tangent vector v is the initial velocity of the curve
a (t) = p Mv, where by Remark 3.5 we are consistently abbreviating v,,
to simply v. Now the image of a under the mapping F is the curve £ such
that
0(0 = F («(*)) = F(p + <v).
We define F*(v) to be the initial velocity 0'(O) of (Fig. 1.16).
Summarizing this process, we obtain the following definition.
7.4 Definition Let F: E" — ► E m be a mapping. If v is a tangent vector to
E" at p, let F*(v) be the initial velocity of the curve t — ► F(p + tv) in E m .
The resulting function F* (from tangent vectors of E" to tangent vectors
of E m ) is called the derivative map F* of F.
Note that the initial position t = of the curve t — > F(p + t\) is F(p).
Thus by Definition 4.3, the point of application of its initial velocity is
F(p). It follows from the definition, then, that F* transforms a tangent
vector to E n at p into a tangent vector to E m at F(p).
For example, let us compute the derivative map of the mapping
F(u,v) = (u  v,2uv)
in (2) of Example 7.3. For a tangent vector v at p, we have
p + tv = (pi + tv u pi + tvt) ;
thus
F(p + tv) = ((pi + tv x f  (p, + tv 2 )\ 2(pi + toi)(pt + tot)).
As t varies, this formula describes that curve in E 2 which, by definition,
F»(v)=/3'(0), z
F ;/l^)f>t*F(T>+tv)
" /F(P)=0(O)
(m = n = 3)
FIG. 1.16
36 CALCULUS ON EUCLIDEAN SPACE [Chap. I
has initial velocity F*(v). Differentiating the coordinates above with
respect to t (Definition 4.3), we obtain.
F*(v) = F(p + tv)'(0) = 2(pi»i  P2V2, vvpi + P1V2) at F(p).
7.5 Theorem Let F = (/i,/ 2 , • • • ,/») be a mapping from E n to E m .
If v is a tangent vector to E" at p, then
F*(v) = (v[/i], ••• ,y[fm]) atF(p).
Thus F* (v) is determined by the derivatives v[f<] 0/ ^e coordinate functions
of F with respect to v.
Proof. For the sake of concreteness we take m = 3. Given v at p, we
refer to the definition (7.4) of F* and let ft be the curve
Pit) =F(p + tv) = (/i(p + tv),Mp + *v),/ 3 (p + tv)).
By definition, /8'(0) = F*(v). According to Definition 4.3, to get the
velocity vector S'(0) we must take the derivatives at t = of the coordi
nate functions /i(p + tv) of 0. But (d/dt)(Ji(p + tv))  ( = is precisely
v[/i], where as usual the point of application p is now omitted from the
notation. Thus
F*(y) = Wi], v[/ 2 ], v[/ 3 ]W
But by definition of /?,
0(0) = F(p). I
Fix a particular point p in E". As noted above, each tangent vector v
to E n at p is transformed by F* into a tangent vector F* (v) to E m at F(p).
Thus for each point p in E", the derivative map F* gives rise to a function
F* p :T p (E n )+T F(p) (E m )
which we call derivative map of F at p. Compare the corresponding situa
tion in elementary calculus where a differentiable function /: R — > R has
a derivative function / : R — > R which at each point t of R gives the deriva
tive /(0 of /at t.
The links between calculus and linear algebra are tighter than one might
expect from a conventional calculus course. A most significant link is
provided by
7.6 Corollary Let F be a mapping from E" to E m . Then at each point
p of E", the derivative map F* p : T p (E n ) — > T F(P )(E m ) is a linear trans
formation.
Proof. If v and w are tangent vectors at p, and a and b are numbers,
we must show that
F*(av + 6w) = aF*(v) + 6F*(w).
Sec. 7] MAPPINGS 37
Using the first assertion in Theorem 3.3, this follows easily from the pre
ceding theorem. 
The linearity of F* p is a generalization of the fact that the derivative
/ (t) of /: R — ► R is the slope of the tangent line to the graph of / at t.
Indeed for each point p, F* p is the linear transformation which best approxi
mates the behavior of F near p. This idea is fully developed in advanced
calculus, where it is used to prove Theorem 7.10.
Since F* p : T p (E n ) —> T F ( P) (E m ) is a linear transformation, it is reasona
ble to compute its matrix with respect to the natural bases
tt(p), ••• , tf.(p) forr p (E n )
CMF(p)), ••• , U m (F(p)) for T np) (E m ).
This matrix is called the Jacobian matrix of F at p.
7.7 Corollary If F = (/i, • • • , f m ) is a mapping from E n to E TO , then
F*(Uj(p)) = £ p. (p) Ui(F(p)) (1 < j < n).
ti ax.
Hence the Jacobian matrix of F at p is ((dfi/dXj) (p))i^i^ m , ig>gn
Proof. Set v = Uj(p) in Corollary 7.6. Since the unit vector Uj(p)
applied to /» is just (dfi/dXj) (p), we get
'• (p ' ( >»  (I <p>> • w f <«•>) § I <*> "'<««■»• ■
Standard abbreviation:
f*(Uj) =T,¥ 1 u i ,
i OjCj
where Uj and dfi/dxj are evaluated at p, and V, is evaluated at F(p).
This result shows that the derivative map of F is completely determined
by the partial derivatives of its coordinate functions. For example, con
sider the second mapping in Example 7.3. Its coordinate functions are
f = u — v and g = 2uv. Hence
2m 2v
2v 2w,
Thus the Jacobian matrix of this mapping at the point p = (pi, P2) is
'2 Pl 2p 2 \
2p 2 2pi
38 CALCULUS ON EUCLIDEAN SPACE [Chap. I
7.8 Theorem Let F: E n * E TO be a mapping. If = F(a) is the image
in E m of the curve a in E", then /?' = F* («').
This theorem asserts that F* preserves velocities of curves, since for each
t, the velocity £ (t) of the image curve is the image, under F*, of the velocity
a (t) of a.
Proof. For definiteness, set m = 3. Now if F = (/ lf / 2 , / 3 ), then
= F(a) = (/x(a),/,(a),/,(a)).
Thus the coordinate functions of are /Si = /,(«). By Theorem 7.5,
f.(«'(0) = («'(0 l/il, «'«) [/ 2 ], «'(0 [/,]).
But applying Lemma 4.6, we find that
„'«) [/.] = «&£!>> (,) = *' (o.
at at
Hence
Furthermore, this tangent vector has point of application f (a(0) = /3(0;
hence it is precisely 3'(<). 
Just as one uses the derivative of a function /: R — ► R to gain informa
tion about the function /, one can use the derivative map F* in the study
of a mapping F. A detailed investigation of this matter belongs in advanced
calculus; we shall give only one or two basic definitions needed in later
work.
7.9 Definition A mapping F: E ra — > E m is regular provided that for each
point p of E" the derivative map F* p is onetoone.
Since each F* p is a linear transformation, we can apply standard results
of linear algebra to conclude that the following conditions are equivalent:
(1 ) F* p is onetoone.
(2) If F* (v„) = 0, then v p = 0.
(3 ) The Jacobian matrix of F at p has rank n (dimension of the domain
E n of F).
For example, the second mapping in Example 7.3 is not regular. But
the onetoone condition fails at only a single point, the origin. In fact,
the computation immediately preceding Theorem 7.8 shows that its
Jacobian matrix has rank 2 at p ?* 0, rank at 0.
A mapping that has an inverse mapping is called a diffeomorphism. A
diffeomorphism is thus necessarily both onetoone and onto, but a mapping
Sec. 7] MAPPINGS 39
which is onetoone and onto need not be a diffeomorphism (Exercise 11).
The results of this section apply equally well to mappings denned only on
open sets of E n . In particular, we may speak of a diffeomorphism from one
open set of E n to another.
We state, without proof, one of the basic results of advanced calculus.
7.1 Theorem Let F: E n — > E n be a mapping such that F* p is onetoone
at some point p. Then there is an open set 11 containing p such that the
restriction of F to 1L is a diffeomorphism 11 — » V onto an open set V.
This is called the inverse function theorem, because it asserts that the
restricted mapping 01 — » V has an inverse mapping V — > 11. The proof is
based on the idea that at points p + Ap very near p, F (p + Ap) is approxi
mately F(p) + F*(Ap). Since the tangent spaces at p and F(p) have
the same dimension, it follows that the onetoone linear transformation
F* p has an inverse; hence so does F — near p.
EXERCISES
1. If F is the mapping F = (u 2 — v 2 , 2uv) in Example 7.3, find all points
p such that
(a) F(p) = (0,0). (b) F(p) = (8, 6). (c) F(p) = p.
2. The mapping F in Exercise 1 carries the horizontal line v = 1 to the
parabola u — > F(u,l) = (u 2 — 1, 2u). Sketch the lines u = 1 and
v = 1, and their images under F.
3. The image F(S) of a set S under a mapping F consists of all points
F(p) with p in S. For F as in Ex. 1, find the image of each of the fol
lowing sets:
(a) The horizontal strip S: 1 ^ v ^ 2.
(b) The half disc S: u 2 + v 2 ^ 1, v ^ 0.
(c) The wedge S: —u ^ v ^ u, u ^ 0.
In each case, show the set S and its image F(S) on a single sketch.
(Hint: Begin by finding the image of the boundary curves of S.)
4. (a) Show that the derivative map of the mapping (1 ) in Example 7.3 is
given by
F*(v P ) = (*>i  t> 2 , v i + *>2, 2v 3 )f( P ).
(Hint: Work directly from the definition of derivative map.)
(b) In general, if F: E n — » E TO is a linear transformation, prove that
F*(v p ) = F(y) F(p) .
40 CALCULUS ON EUCLIDEAN SPACE [Chap. I
5. If F = (f h ••• ,f m ) is a mapping from E n to E m , we write
F* = (d/i, • • • , <*/„,),
since by Theorem 7.5,
F*M = (dfi(\ p ), ■■■ ,df m (v p )) Hp) .
Find F* for the mapping F = (x cos ?/, z sin y, z) from E 3 to E 3 , and
compute F* (v p ) if
(a) v= (2,1,3), p = (0,0,0).
(b) v= (2,1,3), p= (2,t/2,»).
6. Is the mapping in the preceding exercise regular?
7. Let F = (/i,/ 2 ) and G = (gi,g?) be mappings from E 2 to E 2 . Compute
the Euclidean coordinate functions of the composite function GF:
E — > E and show that it is a mapping.
8. In the definition (7.4) of F*(v p ), show that the straight line may be
replaced by any curve a with initial velocity v p .
9. Prove that a mapping F: E n — > E TO preserves directional derivatives
in this sense: If v p is a tangent vector to E n and g is a differentiate
function on E m , then F* (v p )\g] = \ p \g(F)].
10. Let F = (/i,/ 2 ) be a mapping from E 2 to E 2 . If for every point q of E 2
the equations
fffi = fi(Pu P*)
have a unique solution
fpi = giiquQ*)
[P2 = g2(qi, q 2 )
prove that F is onetoone and onto, and that F~ 1 = (^1,^2).
11. (Continuation). In each case, show that F is onetoone and onto,
compute the inverse function F 1 , and decide whether F is a diffeo
morphism (that is, whether F 1 is differentiate).
(a) F = (ve u , u).
(b) F = (u 3 ,v  u).
(c) F = (1 + 2w  2v, 4  2u + 0).
12. Let F; E n ► E m and G: E m > E p be mappings.
(a) Generalize the results of Exercise 7 to this case.
(b) If a is a curve in E n )show that (£F)* («') = (?* (F* («') ). [tfmfc
(GF)(a) = G(F( a )).]
Sec. 8] SUMMARY 41
(c) Deduce that (GF)* = G*F* : The derivative map of a composi
tion of mappings is the composition of their derivative maps.
13. If /: R — ► R is a differentiable realvalued function on the real line
R, prove that /* (v p ) is the tangent vector /'(p) v at the point /(p).
8 Summary
Starting from the familiar notion of realvalued functions, and using linear
algebra at every stage, we have constructed a variety of mathematical
objects. The basic notion of tangent vector led to vector fields, which
dualized to 1 forms— which in turn led to arbitrary differential forms. The
notions of curve and differentiable function were generalized to that of a
mapping F: E" — * E m .
Then starting from the usual notion of the derivative of a realvalued
function, we proceeded to construct appropriate differentiation operations
for these objects: the directional derivative of a function, the exterior
derivative of a form, the velocity of a curve, the derivative map of a
mapping. These operations all reduced to (ordinary or partial) derivatives
of realvalued coordinate functions, but it is noteworthy that in most cases
the definitions of these operations did not involve coordinates. (This could
be achieved in all cases.) Generally speaking, the differentiation operations
all exhibited in one form or another the characteristic linear and Leibnizian
properties of ordinary differentiation.
Most of these concepts are probably already familiar to the reader, at
least in special cases. But we now have careful definitions and a catalogue
of basic properties which will enable us to begin our exploration of differ
ential geometry.
CHAPTER
II
Frame Fields
Roughly speaking, geometry begins with the measurement of distances
and angles. We shall see that the geometry of Euclidean space can be
derived from the dot product, the natural inner product on Euclidean space.
Much of this chapter is devoted to the geometry of curves in E . We
emphasize this topic not only because of its intrinsic importance, but also
because the basic method used to investigate curves has proved effective
throughout differential geometry. A curve in E is studied by assigning at
each point a certain frame — that is, set of three orthogonal unit vectors.
The rate of change of these vectors along the curve is then expressed in
terms of the vectors themselves by the celebrated Frenet formulas (Theorem
3.2). In a real sense the theory of curves inE 3 is merely a corollary of these
fundamental formulas.
Later on we shall use this "method of moving frames" to study a surface
in E 3 . The general idea is to think of a surface as a kind of twodimensional
curve and follow the Frenet approach as closely as possible. To carry out
this scheme we shall need the generalization (Theorem 7.2) of the Frenet
formulas devised by E. Cartan. It was Cartan who, at the beginning of this
century, first realized the full power of this method not only in differential
geometry but also in a variety of related fields.
1 Dot Product
We begin by reviewing some basic facts about the natural inner product
on the vector space E .
1.1 Definition The dot product of points p = (pi, />2, P3) and q = (qi, q 2 , (ft)
in E is the number
p«q = piqi + ZMa + Pzq$
42
Sec. 1]
DOT PRODUCT
43
The dot product is an inner product, that is, it has three properties
(1) Bilinearity:
(ap + 6q)»r = ap»r + 6q«r
r« (ap + 6q) = ar»p + &r»q.
(2) Symmetry: p«q = q»p.
(3) Positive definiteness:p*p ^ 0, andp«p = if and only if p = 0.
(Here p, q, and r are arbitrary points of E 3 , and a and b are numbers. )
The norm of a point p = (pi, p 2 , Pz) is the number
Up 11 = (pp) 1/2 = (pi + vi + Pa 2 ) 1/2 .
The norm is thus a realvalued function on E 3 ; it has the fundamental
properties  p + q  ^  p  +  q  and [ op  =  a  p , where  a  is
the absolute value of the number a.
In terms of the norm we get a compact version of the usual distance
formula in E 3 .
1.2 Definition If p and q are points of E 3 , the Euclidean distance from
p to q is the number
d(p,q) = p  q II
In fact, since
p  q = (px  q u pi  q 2 , p z  q 3 ),
expansion of the norm gives the wellknown formula (Fig. 2.1)
d(p,q) = ((Pi ~ <fc) 2 + (P*  <? 2 ) 2 + (p,  qzf) m .
Euclidean distance may be used to give a more precise definition of open
sets (Chapter 1, Sectionl ). First, if p is a point of E 3 and e > is a number,
the tneighhorhood 9l t of p in E 3 is the set of all points q of E 3 such that
d(p, q) < e. Then a subset of E 3 is open provided that each point of
has an eneighborhood which is entirely contained in 0. In short, all points
near enough to a point of an open set are also in the set. This definition is
X 7
Pi — Qs
Pi — ?2
FIG. 2.1
44
FRAME FIELDS
[Chap. II
valid with E replaced by E" — or indeed any set furnished with a reasonable
distance function.
We saw in Chapter I that for each point p of E 3 there is a canonical iso
morphism v — ► v p from E 3 onto the tangent space T P (E 3 ) at p. These iso
morphisms lie at the heart of Euclidean geometry — using them, the dot
product on E itself may be transferred to each of its tangent spaces.
1.3 Definition The dot product of tangent vectors v p and w p at the
same point of E 3 is the number v p «w p = vw.
For example, (1,0, 1)„.(3, 3,7), = 1(3) + 0(3) + (1)7 = 4.
Evidently this definition provides a dot product on each tangent space
T P (E ) with the same properties as the original dot product on E 3 . In
particular, each tangent vector \ p toE 3 has norm (or length)  \ p  = [ v .
A fundamental result of linear algebra is the Schwarz inequality
 vw  ^  v   w  . This permits us to define the cosine of the angle
# between v and w by the equation (Fig. 2.2).
vw =
W COS d.
Thus the dot product of two vectors is the product of their lengths times
the cosine of the angle between them. (The angle # is not uniquely deter
mined unless further restrictions are imposed, say ^ & ^ x.)
In particular, if & = r /2, then vw = 0. Thus we shall define two vec
tors to be orthogonal provided their dot product is zero. A vector of length
1 is called a unit vector.
1.4 Definition A set d, e2 , e 3 of three mutually orthogonal unit vectors,
tangent to E at p, is called a frame at the point p.
Thus d, e 2 , e 3 is a frame if and only if
ei»ei = d'd = e 3 »e 3 = 1
ei»e 2 = ei»e 3 = e2»e 3 = 0.
v • e 3 e 3
v • e 2 e 2
Sec. 1] DOT PRODUCT 45
By the symmetry of the dot product, the second row of equations is, of
course, the same as
e2*ei = e 3 «ei = e 3 »e2 = 0.
Using index notation, all nine equations may be concisely expressed as
e,e, = 8ij for 1 ^ i,j ^ 3, where 5 tJ is the Kronecker delta (0 if i 9± j,
1 if i = j). For example, at each point p of E , the vectors t/i(p), ?7 2 (p),
C7 3 (p) of Definition 2.4 in Chapter I constitute a frame at p.
1.5 Theorem Let ei, 62, e 3 be a frame at a point p of E . If v is any tan
gent vector to E 3 at p, then (Fig. 2.3)
v = (vei)ei + (ve2)e2 + (ve 3 )e 3 .
Proof. First we show that the vectors e x , d, e 3 are linearly independent.
Suppose
22 «;«; = 0.
Then
= (22 a&i)'ej = 22 a,e,ej = 22 a * 5 *7 — a h
where all sums are over i = 1, 2, 3. Thus
Q>i =■ (h = a 3 = 0,
as required. Now the tangent space T p (E ) has dimension 3, since it is
linearly isomorphic to E 3 . Thus by a wellknown theorem of linear algebra,
the three independent vectors ei, e^, e 3 form a basis for T p (E 3 ). Hence for
each vector v there are three (unique) numbers ci, C2, c 3 such that
v = 22 C &i
But
v*e, = (22 c **i)' e j = 22 c &a = °h
and thus
v = 22 (veOc,. 
This result (valid in any innerproduct space) is one of the great labor
saving devices in mathematics. For to find the coordinates of a vector v
with respect to an arbitrary basis, one must in general solve a set of
nonhomogeneous linear equations, a task which even in dimension 3 is
not always entirely trivial. But the theorem shows that to find the co
ordinates of v with respect to a frame (that is, an orthonormal basis) it
suffices merely to compute the three dot products vei, ve2, ve 3 . We call
this process orthonormal expansion of v in terms of the frame ei, e%, e 3 .
46 FRAME FIELDS [Chap. II
In the special case of the natural frame Ui(p), £/ 2 (p), ^(p) the identity
is an orthonormal expansion, and the dot product is denned in terms of
these Euclidean coordinates by vw = 52 VxW% If we use instead an arbitrary
frame ej, 62, e 3 , then each vector v has new coordinates a* = ve» relative
to this frame, but the dot product is still given by the same simple formula
v«w = 52 ajbi
since
vw = (52 o<e.)»(22 &***■) = S ai ^ j e *" e '
When applied to more complicated geometric situations, the advantage
of using frames becomes enormous, and this is why they appear so fre
quently throughout this book.
The notion of frame is very close to that of orthogonal matrix.
1 .6 Definition Let ei, eg, e 3 be a frame at a point p of E . The 3X3 matrix
A whose rows are the Euclidean coordinates of these three vectors is called
the attitude matrix of the frame.
Explicitly, if
©1 = ( a llj a 12, 0,\z)p
€2 = (fl21, «22, Ovijp
e% = (031, «32, azz) P
then
Thus A does describe the "attitude" of the frame in E , although not its
point of application.
Evidently the rows of A are orthonormal, since
52* a>ikQ>jk = ei»e, = 8ij for 1 = i,j ^ 3.
By definition, this means that A is an orthogonal matrix.
In terms of matrix multiplication, these equations may be written
Sec. 1]
DOT PRODUCT
47
A *A = /, where / is the 3 X 3 identity matrix, and l A is the transpose
oiA:
(an «2i «3i^
ai2 022 a,zz
a U Ct23 «33/
It follows, by a standard theorem of linear algebra, that l AA = I, so that
l A = A 1 , the inverse of A.
There is another product on E 3 , closely related to the wedge product of
1 forms, and second in importance only to the dot product. We shall trans
fer it immediately to each tangent space of E .
1.7 Definition If v and w are tangent vectors to E at the same point p,
then the cross product of v and w is the tangent vector
Ui(p) U t (p) U»(p)
vXw= fi v 2 v 3
Wi V)2 Wz
This formal determinant is to be expanded along its first row. For ex
ample, if v = (1,0, l) p and w = (2,2, 7)„ then
v X w =
£A(p) U t (p) Ut(p)
1 1
2 27
= 2*7i(p) + 5t/ 2 (p) + 2t/,(p) = (2,5,2),.
Using familiar properties of determinants, we see that the cross product
v X w is linear in v and in w, and satisfies the alternation rule
v X w = — w X v
(hence, in particular, v X v = 0). The geometric usefulness of the cross
product is based mostly on
1.8 Lemma The cross product v X w is orthogonal to both v and w,
and has length such that
 v X w  2 = vv ww — (vw) .
Proof. Let v X w = ^dlhip). Then the dot product v(v X w) is
just ^2 v^Ci. But by the definition of cross product, the Euclidean coordinates
Ci, c 2 , Cj of v X w are such that
v» (v X w) =
Vi
V2
v 3
Vi
Vi
v 3
Wi
w 2
w 3
48 FRAME FIELDS
[Chap. II
FIG. 2.4
This determinant is zero, since two of its rows are the same; thus v X w
is orthogonal to v— and, similarly, to w.
Rather than use tricks to prove the length formula, we give a bruteforce
computation. Now
v . vw . w _ (vw) » = (2> 2 )(2>/) _ (j: VlWi) >
= X) vfw*  {£ vlwi + 2 X) ViWiVjWj
= 2J vfwf — 2^2 ViWiVjWj.
i<]
On the other hand,
 v X w 2 = (v X w) v (v Xwj^c, 2
= {ViW Z — ViW 2 f + (V 3 W X — ViW 3 f + (V!W 2 — V 2 Wi) 2
and expanding these squares gives the same result as above. 
A more intuitive description of the length of a cross product is
 v X w  =  v   w  sin 0,
where ^ i? ^ tt is the smaller of the two angles from v to w. The direction
of v X w on the line orthogonal to v and w is given, for practical purposes,
by this "righthand rule": If the fingers of the right hand point in the
direction of the shortest rotation of v to w, then the thumb points
in the direction ofvXw (Fig. 2.4).
Combining the dot and cross product, we get the triple scalar product,
which assigns to any three vectors u, v, w the number u»v X w (Exercise 4).
Parentheses are unnecessary: u(v X w) is the only possible meaning.
EXERCISES
1. Let v = (1,2,1) and w = (  1 , 0, 3 ) be tangent vectors at a point of
E 3 . Compute (a) vw. (b) v X w.
Sec. 1] DOT PRODUCT 49
(c) v/v,w/w. (d) vXw.
(e) the cosine of the angle between v and w.
2. Prove that Euclidean distance has the properties
(a) d(p,q) ^ 0; d(p,q) = if and only if p = q,
(b) d(p,q) = d(q,p),
(c) d(p,q) + d(q,r) ^ d(p,r),
for any points p, q, r in E 3 .
3. Prove that the tangent vectors
(1, 2, 1) (2,0,2) (1,1,1)
61 = V6 ' C2 = VS ' C3 = V3
constitute a frame. Express v= (6, 1, —1) as a linear combination
of these vectors. (Check the result by direct computation. )
4. Let u = (wi, t*2, Us), v = {vi,Vi,Vz), w = (u?i, 102,^3) Prove that
(a) u»v X w =
Wi W2 w 3
Vi v 2 Vz
Wi W2 w z
(b) u»v X w 5^ if and only if u, v, and w are linearly independent.
(c) If any two vectors inu»v X w are reversed, the product changes
sign. Explicitly,
u»v X w = v»w X u = w»u X v
= — wv X u = — vu X w = — u»w X v.
(d) u*v X w = u X v»w.
5. Prove that vXw^Oif and only if v and w are linearly independent,
and show that  v X w  is the area of the parallelogram with sides v
and w.
6. If ei, e 2 , e 3 is a frame, show that
ei»e 2 X e 3 = ±1.
Deduce that any 3X3 orthogonal matrix has determinant ±1.
7. If u is a unit vector, then the component of v in the u direction is
vuu = II v II COS #u.
Show that v has a unique expression v = vi + V2, where Vi«V2 =
and vi is the component of v in the u direction.
8. Show that the volume of the parallelopiped with sides u, v, \r is
±u«v X w (Fig. 2.5). (Hint: Use the indicated unit vector e = v X w/
II v X w .)
50
FRAME FIELDS
[Chap. II
9. Give rigorous proofs, using e
neighborhoods, that each of the
following subsets of E 3 is open:
(a) All points p such that p 
< 1.
(b) All p such that p 3 > 0.
(Hint:  Vi  Qi  ^
d(p,q))
10. In each case, let S be the set of F  G . 2.5
all points p that satisfy the
given condition. Describe S, and decide whether it is open.
(a) V\ + P2 2 + vi = 1 (c) pi = p 2 ^ p 3 .
(b) Pz * 0. ( d ) Pl 2 + P2 2 < 9.
11. If / is a differentiate function on E 3 , show that the gradient
Vf = T,(df/dXi)U i
(Ex. 8 of 1.6) has the following properties:
(a) v[/] = (df ) (v) = v. (V/ ) (p) for any tangent vector at p.
(b) The norm (V/)(p) = [Z(df/dx l )\p)'] m of (V/)(p) is the
maximum of the directional derivatives u[/ ] for all unit vectors at p.
Furthermore, if (V/)(p) ^ the unit vector for which the maximum
occurs is
(V/)(p)/(V/)(p).
The notations grad/, curl V, and div V (in the exercise referred to)
are often replaced by V/, V X V, and V*V, respectively.
12. Angle functions. Let/ and g be differentiable realvalued functions on
an interval /. Suppose that f + g 2 = 1 and that # is a number such
that/(0) = cos #0, g(0) = sin tf . If # is the function such that
Ht) = tfo + / (fg  gf) dt,
prove that
/ = cos t?, g = sin &.
(Hint: We want (/  cos #) 2 + (g  sin i?) 2 = 0; show that
(/cost? + gsm&)' = 0.)
The point of this exercise is that # is a differentiable function, un
ambiguously defined on the whole interval /.
Sec. 2] CURVES 51
2 Curves
We begin the geometric study of curves by reviewing some familiar defini
tions. Let a: I — »E 3 be a curve. In Chapter I, Section 4, we defined the
velocity vector a (t) of a at t. Now we define the speed of a at t to be the
length v(t) =  a' (Oil of the velocity vector. Thus speed is a realvalued
function on the interval /. In terms of Euclidean coordinates a = (cm, 02, "3),
we have
Hence the speed function v of a is given by the usual formula
In physics, the distance traveled by a moving point is determined by
integrating its speed with respect to time. Thus we define the arc length
of a from t = a to t = b to be the number
f«'(OII<a.
Substituting the formula for  a  given above, we get the usual formula
for arc length.
Sometimes one is interested only in the route followed by a curve and
not in the particular speed at which it traverses its route. One way to ig
nore the speed of a curve a is to reparametrize to a curve /3 which has unit
speed  j8'  = 1. Then ft represents a "standard trip" along the route of a.
2.1 Theorem If a is a regular curve in E 3 , then there exists a reparame
trization of a such that /3 has unit speed.
Proof. Fix a number a in the domain / of a: I — > E 3 , and consider the
arclength function
du.
(The resulting reparametrization is said to be based at t = a.) Thus the
derivative ds/dt of the function s = s(t) is the speed function v = \\ a \\
of a. Since a is regular, by definition a is never zero; hence ds/dt > 0.
By a standard theorem of calculus, the function s has an inverse function
t = t(s), whose derivative dt/ds at s = s(t) is the reciprocal of ds/dt at
t = t(s). In particular, dt/ds > 0.
52 FRAME FIELDS
[Chap. II
Now let j8 be the reparametrization 0(a) = a (<(«)) of a. We assert that
has unit speed. In fact, by Lemma 4.5 of Chapter I,
0'(«) = (dt/ds)(s)a(t(s)).
Hence, by the preceding remarks, the speed of is
llis'00  =  (.)  «'(«(«))  =  (s)  s («(,)) = i. i
We shall use the notation of this proof frequently in later work. The
unitspeed curve is sometimes said to have arclength parametrization,
since the arc length of from s = a to s = b (a < b) is just b — a.
For example, consider the helix a in Example 4.2 of Chapter I. Since
a(t) = (a cos t, a sin t, bt), the velocity a is given by the formula
a (t) = ( — o sin t, a cos £, 6).
Hence
 a'COII 1 = a(t)>a'(t) = a 2 sin 2 < + a 2 cos 2 * + 6 2 = a 2 + 6 2 .
Thus a has constant speed c =  a \\ = (a 2 + fe 2 ) 1/2 . If we measure arc
length from t = 0, then
»(0 = f
Jo
c dt = ct.
Hence, t(s) = s/c Substituting in the formula for a, we get the unitspeed
reparametrization
fi(s) = «() = (« cos , a sin  , — J .
It is easy to check directly that  j8'(s) = 1 for all s.
A reparametrization a{h) of a curve a is said to be orientationpreserving
if h > 0, orientationreversing if h < 0. In the latter case, a and a(h) tra
verse their common route in opposite directions. By the conventions above,
a unitspeed reparametrization is always orientationpreserving since
ds/dt > for a regular curve a.
We now define a variant of the general notion of vector field (Definition
2.3 of Chapter I) which is adapted to the study of curves. Roughly speak
ing, a vector field on a curve consists of a vector at each point of the curve.
2.2 Definition A vector field Y on a curve a : / — * E is a function that
assigns to each number t in / a tangent vector Y(t) to E 3 at the point
a(t).
We have already used such vector fields, since for any curve a, its ve
Sec. 2]
CURVES
53
FIG. 2.6
locity a evidently satisfies this definition. Note that, unlike a , arbitrary
vector fields on a need not be tangent to a, but may point in any direction
(Fig. 2.6).
The properties of vector fields on curves are analogous to those of vector
fields on E . For example, if 7 is a vector field on a : I — > E , then for each
t in / we can write
7(0 = (yi(t),y2(t),y 3 (t)) a(t) = £ yS) Ui(a{t)).
We have thus defined realvalued functions y x , y 2 , y* on / called the
Euclidean coordinate junctions of 7. These will always be assumed to be
differentiable. Note that the composite function t — > Ui(a(t)) is a vector
field on a. Where it seems safe to do so, we shall often write merely £/ t
instead of Ui(a(t)).
The operations of addition, scalar multiplication, dot product, and cross
product of vector fields (on the same curve) are all defined in the usual
pointwise fashion. Thus if
7(0 = eu,  tU 3 , Z(t) = (1  t 2 )U 2 + tU 3 ,
and
fit) =
t + 1
t
we obtain the vector fields
(7 + Z)(0 = * 2 t/i+ (1  t 2 )U*
(/7)(0 = <« + l)t/i  (*+ l)C/ 3
t/i
u 2 u z
(7XZ)(0 =
t 2
t
1  1 2 t
= t(l  t 2 )Ui  t 3 U 2 + t 2 (l  t 2 )U 3
54 FRAME FIELDS [Chap. II
and the realvalued function
(YZ)(t) = t\
To differentiate a vector field on a one simply differentiates its Euclidean
coordinate functions, thus obtaining a new vector field on a. Explicitly, if
Y = 53 y%Ui, then Y' = YL (dyi/dt)Ui. Thus, for Y as above, we get
Y' = 2tU,  U 3 , Y" = 2U U and Y'" = 0.
In particular, the derivative a" of the velocity a of a is called the accelera
tion of a. Thus if a = (ai,a2,a 3 ), the acceleration a" is the vector field
a ~ \dt 2 ' W ~d¥/ a
on a. By contrast with velocity, acceleration is generally not tangent to
the curve.
As we mentioned earlier, in whatever form it appears, differentiation
always has suitable linearity and Leibnizian properties. In the case of
vector fields on a curve, it is easy to prove the linearity property
(aY + bZY = aY' + bZ'
(a and b numbers) and the Leibnizian properties
(/F)' = ^ Y + jY' and (YZ)' = Y'Z + YZ'.
If the function Y'Z is constant, the last formula shows that
F'.Z + YZ' = 0.
This observation will be used frequently in later work. In particular, if F
has constant length  F , then F and F are orthogonal at each point,
since  F  2 = Y*Y constant implies 2F»F' = 0.
Recall that tangent vectors are parallel if they have the same vector
parts. We say that a vector field F on a curve is parallel provided all its
(tangent vector) values are parallel. In this case, if the common vector
part is (ci, c 2 , c 3 ), then
Y(t) = (ci,c 2 ,c 3 ) aa) = X dUi for all*.
Thus parallelism for a vector field is equivalent to the constancy of its
Euclidean coordinate functions.
Vanishing of derivatives is always important in calculus; here are three
simple cases.
2.3 Lemma (1) A curve a is constant if and only if its velocity is zero,
a = 0.
Sec. 2] CURVES 55
(2) A nonconstant curve a is a straight line if and only if its acceleration
is zero, a." = 0.
(3) A vector field 7 on a curve is parallel if and only if its derivative
is zero, 7 = 0.
Proof. In each case it suffices to look at the Euclidean coordinate func
tions. For example, we shall prove (2). If a = (ai, as, a 3 ), then
„ _ /(fai (fat2 d 2 aa \
Thus a" = if and only if each (Fai/dt 2 = 0. By elementary calculus, this
is equivalent to the existence of constants p» and #»• such that
cti(t) = pi + tq u for i = 1, 2, 3.
Thus a(t) = p + tq, and a is a straight line as defined in Example 4.2 of
Chapter I. (Note that nonconstancy implies q^O.) 
EXERCISES
1. For the curve a (0 = (2t, f, t 3 /3),
(a) find the velocity, speed, and acceleration for arbitrary t, and at
t = 1;
(b) find the arclength function s = s(t) (based at t = 0), and de
termine the arc length of a from t = — 1 to t = +1.
2. Show that the curve a (t) = (t cos t, t sin t, t) lies on a cone in E 3 . Find
the velocity, speed, and acceleration of a at the vertex of the cone.
3. Show that the curve a(t) = (cosh t, sinh t, t) has arclength function
s (t) = y/2 sinh t, and find a unitspeed reparametrization of a.
4. Consider the curve a(t) = (2t, t\ log t) on /: t > 0. Show that this
curve passes through the points p = (2,1,0) andq = (4, 4, log 2), and
find its arc length between these points.
5. Suppose that ft and ft are unitspeed reparametrizations of the same
curve a. Show that there is a number s such that ft(s) = ft(s + s )
for all s. What is the geometric significance of s ?
6. Let Y be a vector field on the helix a(t) = (cos t, sin t, t). In each of
the following cases, express Y in the form ^Z !/&%'■
(a) Y(t) is the vector from <x(t) to the origin of E 3 .
(b) Y(t) = a'(t) *"(t).
(c) 7(0 has unit length and is orthogonal to both a (t) and a"{t).
(d) 7(0 is the vector from a(t) to a(t + rr).
56 FRAME FIELDS [Chap. II
7. Let F be a vector field on a curve a. If a (h) is a reparametrization of
a, show that F(/&) is a vector field on a(h), and prove the chain rule
Y(h)' = A' F'(/i).
8. Let a,(i: I — »E be curves such that a (t) and /?'(£) are parallel (same
Euclidean coordinates) for each t. Prove that a and are parallel in
the sense that there is a point p in E 3 such that p{t) = a(t) + p for
alH.
9. If a is a regular curve show that
(a) a has constant speed if and only if the acceleration a" is always
orthogonal to a (that is, to a').
(b) a is a reparametrization of a straight line t — » p + tq if and only if
a" is always tangent to a (that is, a" and a are collinear).
10. A portion of a curve defined on a closed interval [a,b]' a ^ t ^ b, is
called a curve segment. A reparametrization a(h): [a,b] — »E 3 of a curve
segment a : [c,d] — > E 3 is monotone provided either
(a) ti ^ 0, h(a) = c, h(b) = d, or (b) ti ^ 0, h(a) = d, h(b) = c.
Prove that monotone reparametrization does not change arc length.
11. Prove that a straight line is the shortest distance between two points
in E . Use the following scheme; let a : [a,b] — > E 3 be an arbitrary curve
segment fromp = a(a) to q = a(b). Let u = (q — p)/q — p .
(a) If a is a straightline segment from p to q, say
a(t) = (1  Op + *q (0 ^ t ^ 1),
show that L(ct) = d(p, q).
(b) From  a  ^ a'«u, deduce L(a) ^ d(p, q), where L(a) is the
length of a and d is Euclidean distance.
(c) Furthermore, show that if L(a) = d(p,q), then (but for para
metrization) a is a straight line segment. (Hint: write a =
(a »u)u + Y, where F*u = 0.)
3 The Frenet Formulas
We now derive mathematical measurements of the turning and twisting of
a curve inE . Throughout this section we deal only with unitspeed curves;
in the next we extend the results to arbitrary regular curves.
Let jS: / — >E be a unitspeed curve, so  j8' (s)  = 1 for each s in /. Then
T = /3 is called the unit tangent vector field on 0. Since T has constant
length 1, its derivative T = $" measures the way the curve is turning
in E . We call T' the curvature vector field of 0. Differentiation of T* T = 1
gives 2T' • T = 0, so T' is always orthogonal to T, that is, normal to /3.
Sec. 3] THE FRENET FORMULAS 57
The length of the curvature vector field T' gives a numerical measure
ment of the turning of (}. The realvalued function k such that k(s) = \\ T (s) \\
for all s in I is called the curvature function of 0. Thus k j^ 0, and the larger
k is, the sharper the turning of 0.
To carry this analysis further, we impose the restriction that k is never
zero, so k > O.f Then the unitvector field N = T'/k on tells the
direction in which /3 is turning at each point. N is called the principal normal
vector field of (Fig. 2.7). The vector field B = T X JV on is then called
the binormal vector field of /?.
3.1 Lemma Let /3 be a unitspeed curve in E 3 with k > 0. Then the three
vector fields T, N, and B on are unit vector fields which are mutually
orthogonal at each point. We call T, N, B the Frenet frame field on /S.
Proof. By definition \\T\\ = 1. Since k =  T'  > 0,
II n  = (iA) ii r  = i.
We saw above that T and N are orthogonal — that is, T'N = 0. Then by
applying Lemma 1.8 at each point, we conclude that  B \\ = 1, and B is
orthogonal to both T and N. I
In summary, we have T = p', N = T'/k, and B = T X N, satisfying
T'T = N'N = BB = 1, with all other dot products zero.
The key to the successful study of the geometry of a curve £ is to use its
Frenet frame field T, N, B whenever possible, instead of the natural frame
field Ui, Ui, U 3 . For the Frenet frame field of £ is full of information about
j9, whereas the natural frame field contains none at all.
The first and most important use of this idea is to express the derivatives
T', N', B' in terms of T, N, B. Since T = #', we have T' = 0" = kN. Next
consider B' . We claim that B' is, at each point, a scalar multiple of N.
To prove this, it suffices by ortho normal expansion to show that B »B =
and B''T = 0. The former holds since B is a unit vector. To prove the
t For an arbitrary unitspeed curve, this means that we must make a separate
study of each segment on which « > 0; see Exercise 19 of Section 4.
58
FRAME FIELDS
[Chap. II
latter, differentiate B*T = 0, obtaining B' *T + BT' = 0; then
B'T = BT' = BkN = 0.
Thus we can now define the torsion function t of the curve to be the
realvalued function on the interval I such that B' = — tN. (The minus
sign is traditional.) By contrast with curvature, there is no restriction on
the values of r — it may be positive, negative, or zero at various points of /.
(Indeed the sign of t, at each point, turns out to have an interesting geo
metric significance.) We shall presently show that t does measure the
torsion, or twisting, of the curve 0.
3.2 Theorem (Frenet formulas) If /?: / — »E 3 is a unitspeed curve with
curvature k > and torsion t, then
T' = kN
N' =  K T + tB.
B' = tN
Proof. As we saw above, the first and third formulas are essentially just
the definitions of curvature and torsion. To prove the second, we use
orthonormal expansion to express N' in terms of T, N, B:
N' = N'T T + N'N N + N'B B.
These coefficients are easily found. Differentiating N'T = 0, we get
N'T + N'T' = 0; hence
N'*T = N^T' = NkN = k.
As usual, N'*N = 0, since N is a, unit vector field. Finally,
N'B = NB' = N(tN) = t. I
3.3 Example We compute the Frenet frame T, N, B and the curva
ture and torsion functions of the unitspeed helix
of \ ( s ■ s bs\
p(s) = o cos  , a sin  , — ) ,
\ c c c/
where c = (a + b 2 ) 112 and a > 0. Now
rrf \ a' f \ ( a . s a s b\
T(s) = (s) = ( sin  ,  cos  ,  )
\ c c c c c)
Hence
T'(s) = ( —  cos , —% sin , 0j.
\ c l c c 2 c /
FIG. 2.8
Thus
Sec. 3] THE FRENET FORMULAS 59
,(.) =  T'M II = « = j^ > 0.
Since T' = nN, we get
iV(s) = ( — cos , — sin , 0)
\ c c }
Note that regardless of what values a and b have, N always points straight
in toward the axis of the cylinder on which /3 lies (Fig. 2.8).
Applying the definition of cross product to B == T X N, we get
t>< \ (b . s b s a\
B{s) = (  sin  , — cos ,  1 .
\c c c c c/
It remains to compute torsion. Now
5 '( s ) = ( 1 cos " > j sin  , J ,
and by definition, B' = — tN. Comparing the formulas for B and N, we
conclude that
( \ b
t(s) =  =
c 2 a 2 + b 2
So the torsion of the helix is also constant.
Note that when the parameter b is zero, the helix reduces to a circle of
radius a. The curvature of this circle is k = l/o (so the smaller the radius,
the larger the curvature), and the torsion is identically zero.
This example is a very special one — in general (as the examples in the
exercises show) neither the curvature nor the torsion functions of a curve
need be constant.
3.4 Remark We have emphasized all along the distinction between a
tangent vector and a point of E 3 . However, Euclidean space has, as we
have seen, the remarkable property that given a point p, there is a natural
onetoone correspondence between points (vi, #2,^3) and tangent vectors
(vi,V2, v 3 ) p ait p. Thus one can transform points into tangent vectors (and
vice versa) by means of this canonical isomorphism. In the next two
sections particularly, it will often be convenient to switch quietly from one
to the other without change of notation. Since corresponding objects have
the same Euclidean coordinates, this switching can have no effect on scalar
multiplication, addition, dot products, differentiation — or any other
operation defined in terms of Euclidean coordinates.
Thus a vector field Y = (^1,2/2,^3)0 on a curve becomes itself a curve
(VuViyVs) in E 3 . In particular, if Y is parallel, its Euclidean coordinate
functions are constant, so Y is identified with a single point of E .
60
FRAME FIELDS
[Chap. II
In solid geometry one describes a plane in E 3 as being composed of all
perpendiculars to a given line at a given point. In vector language then, the
plane through p orthogonal to q ^ consists of all points r in E 3 such that
(r — p)»q = 0. By the remark above, we may picture q as a tangent
vector at p as shown in Fig. 2.9.
We can now give an informative approximation of a given curve near
an arbitrary point on the curve. The goal is to show how curvature and
torsion influence the shape of the curve. To derive this approximation we
use a Taylor approximation of the curve — and express this in terms of the
Frenet frame at the selected point.
For simplicity, we shall consider the unitspeed curve = (ft, ft, ft) near
the point /8(0). For s small, each coordinate ft 6s) is closely approximated
by the initial terms of its Taylor series:
ft(s)
ft(0) + ^ (0)
as
• + &<»£ + £«»•
Hence
0(s) — /3(0) +
*'(0) + /3"(0) +/3'"(0).
But jS'(O) = To, and /3"(0) = koN , where the subscript indicates evalua
tion at s = 0, and we assume k ^ 0. Now
f = UN)' = ^N + kN'.
as
Thus by the Frenet formula for N', we get
j8'"(0) = ko 2 T + ^ (0) N + K ToBo.
as
FIG. 2.9
0(8) ~/J(s)
FIG. 2.10
Sec. 3] THE FRENET FORMULAS 61
Finally, substitute these derivatives into the approximation of 0(s) given
above, and keep only the dominant term in each component (that is, the
one containing the smallest power of s). The result is
2 3
0(s) ~ 0(0) + sT + ko I N + KOTO g #0.
Denoting the right side by 0(s), we obtain a curve called the Frenet
approximation of near s = 0. We emphasize that has a different Frenet
approximation near each of its points; if is replaced by an arbitrary
number s , then s is replaced by s — s , as usual in Taylor expansions.
Let us now examine the Frenet approximation given above. The first
term in the expression for is just the point 0(0). The first two terms
give the tangent line s — ► 0(0) + sT of at 0(0)— the best linear approxi
mation of near 0(0). The first three terms give the parabola
s>/8(0) + sT + ko(s 2 /2)N ,
which is the best quadratic approximation of near 0(0). Note that this
parabola lies in the plane through 0(0) orthogonal to B , the osculating
plane of at 0(0). This parabola has the same shape as the parabola
y = Ko x 2 /2 in the xy plane, and is completely determined by the curvature
ko of at s = 0.
Finally, the torsion r , which appears in the last and smallest term of 0,
controls the motion of orthogonal to its osculating plane at 0(0), as
shown in Fig. 2.10.
On the basis of this discussion, it is a reasonable guess that if a unit
speed curve has curvature identically zero, then it is a straight line. In fact,
this follows immediately from (2) of Lemma 2.3, since k =  T  =  0" ,
so that /t = if and only if 0" = 0. Thus curvature does measure deviation
from straightness.
A plane curve inE 3 is a curve that lies in a single plane of E 3 . Evidently
a plane curve does not twist in as interesting a way as even the simple
helix in Example 3.3. The discussion above shows that for s small the curve
tends to stay in its osculating plane at 0(0); it is t j^ which causes
to twist out of the osculating plane. Thus if the torsion of is identically
zero, we may well suspect that never leaves this plane.
3.5 Corollary Let be a unitspeed curve in E 3 with k > 0. Then
is a plane curve if and only if t = 0.
Proof. Suppose is a plane curve. Then by the remarks above, there
exist points p and q such that (0(s) — p)*q = for all s. Differentiation
yields
0' (s) q = 0" («) q = for all s.
62
[Chap. II
FIG. 2.11
Thusq is always orthogonal to T = /?' and N = 0"/ K . But B is also orthog
onal to T and N, so, since B has unit length, B = ±q/ q . Thus B f = 0,
and by definition t = (Fig. 2.11).
Conversely, suppose r = 0. Thus B' = 0; that is, B is parallel and may
thus be identified (by Remark 3.4) with a point of E 3 . We assert that
lies in the plane through (3(0) orthogonal to B. To prove this, consider the
realvalued function
/(*) = (0(«) ~ (()))•£ for alls.
Then
df
ds
/ = B = TB = 0.
But obviously, /(0) = 0, so /is identically zero. Thus
(0(s)  i8(0)).5 = for alls,
which shows that lies entirely in this plane orthogonal to the (parallel)
binormal of /3. ■
We saw at the end of Example 3.3 that a circle of radius a has curvature
1/a and torsion zero. Furthermore the formula given there for the principal
normal shows that for a circle, N always points toward its center. This sug
gests how to prove the following converse.
3.6 Lemma If is a unitspeed curve with constant curvature k >
and torsion zero, then is part of a circle of radius \/k.
Proof. Since r = 0, is a plane curve. What we must now show is that
every point of is at distance \/k from some fixed point— which will thus
be the center of the circle. Consider the curve 7 = + (1/k)N. Using the
hypothesis on 0, and (as usual) a Frenet formula, we find
0' +±N' = T +{ K T) = 0.
Sec. 3]
THE FRENET FORMULAS
63
Hence the curve y is constant; that is, /3(s) + (1/k)N(s)
has the same value, say c, for all s (see Fig. 2.12). But
the distance from c to /3(s) is
d(c, (S(s)) = c/8(«) =
 N(s)
K
I
In principle, every geometric problem about curves fig. 2.12
can be solved by means of the Frenet formulas. In simple
cases it may be just enough to record the data of the problem in con
venient form, differentiate, and use the Frenet formulas. For example,
suppose (3 is a unitspeed curve that lies entirely in the sphere 2 of ra
dius a centered at the origin of E 3 . To stay in the sphere, must curve; in
fact it is a reasonable guess that the minimum possible curvature occurs
when ^ is on a great circle of 2. Such a circle has radius a, so we con
jecture that a spherical curve /3 has curvature a ^ 1/a, where a is the radius
of its sphere.
To prove this, observe that since every point of 2 has distance a from
the origin, we have /3«/3 = a 2 . Differentiation yields 2/3' •/? = 0, that is,
fi*T = 0. Another differentiation gives p'*T + P'T' = 0, and by using a
Frenet formula we get T»T + k(}*N = 0; hence
K p.N = 1.
By the Schwarz inequality,
I fi.N I ^    N  = a,
and since k ^ Owe obtain the required result:
(3N
Continuation of this procedure leads to a necessary and sufficient condi
tion (expressed in terms of curvature and torsion) for a curve to be spheri
cal, that is, lie on some sphere in E 3 (Exercise 10).
EXERCISES
1. Compute the Frenet apparatus k, t, T, N, B of the unitspeed curve
/8(s) = (f cos s, 1 — sin s,
find its center and radius.
cos s). Show that this curve is a circle;
2. Consider the curve
0(s)
("■
+ S) 3 ' 2 (18)
3 ' 3
3/2
'V2)
64
FRAME FIELDS
[Chap. II
defined on I: —1 < s < 1. Show that has unit speed, and compute
its Frenet apparatus.
3. For the helix in Example 3.3, check the Frenet formulas by direct
substitution of the computed values of k, t, T, N, B.
4. Prove that
T=NXB = BXN
N = BXT=TXB
B = T X N = N X T.
(A formal proof uses properties of the cross product established in
the Exercises of Section 1 — but one can recall these formulas by using
the righthand rule given on p. 48.)
5. If A is the vector field tT + kB on a unitspeed curve 0, show that
the Frenet formulas become
T' = A X T
N' = A XN
B' = A XB.
6. A unitspeed parametrization of a circle may be written
y(s) = c + r cos s/r e x + r sin s/r e 2 ,
where e^e, = 5^.
If is a unitspeed curve with k(0) > 0, prove that there is one
and only one circle y which approximates £ near #(0) in the sense
that
7(0) = 0(0), t'(0) = /?'(0), and 7 "(0) = /?"(0).
Show that 7 lies in the osculating plane of /3 at /3(0) and find its center
c and radius r. The circle y is called the osculating circle and c the
0(0) = 7(0)
FIG. 2.13
FIG. 2.14
Sec. 3] THE FRENET FORMULAS 65
center of curvature of (3 at /3(0). (The same results hold when is re
placed by any number s. )
7. If a and a reparametrization a = a(h) are both unitspeed curves,
show that
(a) h(s) = dts + So for some number s
(b) f = ± T(h)
N = N(h) H = K (h) f = r(h)
B = ±B(h)
where the sign (db) is the same as that in (a), and we assume k >
(Fig. 2.14). Thus even in the orientationreversing case, the principal
normals N and N still point in the same direction.
8. Curves in the plane. For a unitspeed curve /3(s) = (x(s), y(s)) inE 2 ,
the unit tangent is T = j8' = {x , y), but the unit normal N is denned
by rotating T through +90°, so N = (y,x). Thus T' and N are
collinear, and the curvature of /3 is denned by the Frenet equation
T' = kN. Prove
(a) N' = kT.
(b) If <p is the slope angle f of /3, then k = <p .
This procedure differs from that forE 3 , since k need not be positive —
indeed its sign tells which way /3 is turning. Furthermore, N is denned
without assuming k > 0.
9. Let /3 be the Frenet approximation of an arbitrary unitspeed curve
near s = 0. If, say, the B component of /3 is removed, the resulting
curve is the orthogonal projection of /3 in the ToNq plane. It is the
view of j8 ~ (8 one gets by looking toward /3(0) = /3(0) directly along
the vector B . Sketch the general shape of the orthogonal projections
of $ on each of the planes T N , T B , N B , assuming t > 0. (These
views of P may be confirmed experimentally using a bent piece of
wire. )
10. Spherical curves. Let a be a unitspeed curve with k > 0, t ^ 0.
(a) If a lies on a sphere of center c and radius r, show that
a — c = — pN — p aB,
where p = 1/k and <r = 1/t. Thus r 2 = p + (p a) 2 .
(b) Conversely, if p f (per) 2 has constant value r and p 7^ 0,
show that a lies on a sphere of radius r.
{Hint: For (b), show that the "center curve" 7 = a + pN + p <r.B —
suggested by (a) — is constant.)
t The existence of <p as a differentiable function with T = cos ^> C/i + sin^> ^de
rives from Exercise 12 of Section 1.
66
FRAME FIELDS
[Chap. II
11. Let /3, $: I — >E be unitspeed curves with nonvanishing curvature and
torsion. If T = f , then and are parallel (Ex. 8 of II.2). If B = B,
prove that /8 is parallel to either f} or the curve s — > —fi(s).
4 ArbitrarySpeed Curves
It is a simple matter to adapt the results of the previous section to the
study of a regular curve a: I — > E 3 which does not necessarily have unit
speed. We merely transfer to a the Frenet apparatus of a unitspeed re
parametrization a of a. Explicitly, if s is an arclength function for a
as in Theorem 2.1, then
a(t) = a(s(t)) for alii
or, in functional notation, a = a(s). Now if H > 0, f, t , N, and B are
denned for a as in Section 3, we define for a the
Curvature function: k = ic(s)
Torsion function : t = f(s)
Unit tangent vector field : T = f(s)
Principal normal vector field: N = N(s)
Binormal vector field : B = B(s)
In general k and k are different functions, defined on different intervals.
But they give exactly the same description of the turning of the common route
of a and a, since at any point a(t) = a(s(t)) the numbers n(t) and k(s(0)
are by definition the same. Similarly with the rest of the Frenet apparatus;
since only a change of parametrization is involved, its fundamental geo
metric meaning is the same as before. In particular, T, N, B is again a frame
field on a linked to the shape of a as indicated in the discussion of Frenet
approximations.
(— •
FIG. 2.15
Sec. 4] ARBITRARYSPEED CURVES 67
For purely theoretical work this simple transference is often all that is
needed. Data about a converts into data about the unitspeed reparametri
zation a; results about a convert to results about a. For example, if a is
a regular curve with t = 0, then by the definition above 5 has f = 0;
by Corollary 3.5, a is a plane curve, so obviously a is, too.
However, for explicit numerical computations — and occasionally for the
theory as well — this transference is impractical, since it is rarely possible
to find explicit formulas for a. (For example, try to find a unitspeed para
metrization for the curve a(t) = (t,f,t).)
The Frenet formulas are valid only for unitspeed curves; they tell
the rate of change of the frame field T, N, B with respect to arc length.
However, the speed v of the curve is the proper correction factor in the
general case.
4.1 Lemma If a is a regular curve in E with k > 0, then
T' = kvN
N' = kvT + tvB.
B' = tvN
Proof. Let a be a unitspeed reparametrization of a. Then by definition,
T = f(s), where s is an arclength function for a. The chain rule as applied
to differentiation of vector fields (Exercise 7 of Section 2) gives
T' = f'(s) ^.
at
By the usual Frenet equations, f = UN . Substituting the function s in
this equation yields
T'(s) = k(s)N(s) = kN
by the definition of k and N in the arbitrary speed case. Since ds/dt is the
speed function v of a, these two equations combine to yield T = kvN.
The formulas for iV"' and B' are derived in the same way. 
There is a commonly used notation for the calculus that completely
ignores change of parametrization. For example, the same letter would
designate both a curve a and its unitspeed parametrization a, and simi
larly with the Frenet apparatus of these two curves. Differences in deriva
tives are handled by writing, say, dT/dt for T , but dT/ds for either f
or its reparametrization f (s). With these conventions, the proof above
would combine the chain rule dT/dt = (dT/ds) (ds/dt) and the Frenet
formula dT/ds = kN to give dT/dt = kvN.
Only for a constantspeed curve is acceleration orthogonal to velocity,
since /3 »/3 constant is equivalent to (/3 •/? ) = 2/3 «jS" = 0. In the general
68
[Chap. It
FIG. 2.16
case, we analyze velocity and acceleration by expressing them in terms of
the Frenet frame field.
4.2 Lemma If a is a regular curve with speed function v, then the ve
locity and acceleration of a are given by (Fig. 2.16)
a = vT a" = fT + kv 2 N
at
Proof. Since a = a(s), where s is the arclength function of a, we find,
using Lemma 4.5 of Chapter I that
a' = oc'(s) ~ = vf(s) = vT.
at
Then a second differentiation yields
at at
where we use Lemma 4.1. I
The formula a = vT is to be expected — a and T are each tangent to
the curve, and T has a unit length while  a \\ = v. The formula for ac
celeration is more interesting. By definition, a" is the rate of change of the
velocity a , and in general both the length and the direction of a are chang
ing. The tangential component (dv/dt)T of a" measures the rate of change
of the length of a (that is, of the speed of a). The normal component kv 2 N
measures the rate of change of the direction of a. Newton's laws of motion
show that these components may be experienced as forces. For example,
in a car that is speeding up or slowing down on a straight road, the only
force one feels is due to (dv/dt)T. If one takes an unbanked curve at speed
v, the sideways force one feels is due to kv 2 N. Here k measures how sharply
the road turns; the effect of speed is given by v 2 , so 60 miles per hour is
four times as unsettling as 30.
Sec. 4] ARBITRARYSPEED CURVES 69
We now find effectively computable expressions for the Frenet appara
tus.
4.3 Theorem Let a be a regular curve in E . Then
T =«7ll«'
N = BXT k =  a X a" H/ll cl  3
B = a X a VII <*' X a"  r = (a X a" )•<*'"/ II «' X a"  2 .
Proof. Since v = \\ a \\ > 0, the formula T = a'/\\ a  is equivalent to
a = vT. From the preceding lemma we get
a X a" = {vT) x(f t T + kV 2 N\
= v ^ T X T + *v 3 r X iV = /a> 3 £
since T X T = 0. Taking norms we find
 a X a"  =  kv 3 B  = kv 3
because  B \\ = 1, k ^ 0, and v > 0. Indeed this equation shows that for
regular curves, \\ a X a."  > is equivalent to the usual condition k > 0.
(Thus for k > 0, a and a" are linearly independent and determine the
osculating plane at each point, as do T and N. ) Then
D a' X a" a' X a"
B = 3
kv s ~  a X «"  '
Now AT = B X T is true for any Frenet frame field (Exercise 4 of Section
3); thus only the formula for torsion remains to be proved. To find the
dot product (a X a" )•<*'" we express everything in terms of T, N, B.
We already know that a X a" = wB. Thus, since = T*B = N'B,
we need only find the B component of a". But
«'" = (■£ T + kv 2 n\ = KV 2 N f +
= kv\B + • • •
where we use Lemma 4.1. Consequently (a X a")*a" = kVt, and since
 a X a."  = kv , we have the required formula for r. 
The triple scalar product in this formula for t could (by Exercise 4
of Section 1) also be written a a" X a". But we need a X a." anyway,
so it is usually easier to find (a X a")* a'".
AA Example We compute the Frenet apparatus of the curve
ait) = cm  t 3 ,3t 2 ,:u + t 3 ).
70
FRAME FIELDS
[Chap. II
The derivatives are
Now
a(t) = 3(1  t\ 2t, 1 + t 2 )
a"(t) = 6(«, 1,0
«'"(0 = 6(l,0,l).
a(t)»a(t) = 18(1 + 2f + t 4 ),
SO
»«) =  a' (Oil = VT&Q. + ?).
Applying the definition of cross product, we find
a'(0 X a" {t) = 18
1  t 2 2t 1 + t 2
t 1 «
= 18 (1 + t 2 , 2t,l + t 2 ).
Dotting this vector with itself, we get
(18) 2 { (1 + t 2 ) 2 + U 2 + (1 + t 2 ) 2 ) = 2(18) 2 (1 + * 2 ) 2 .
Hence
 a (0 Xa"(OI = 18V2 (1 + i 2 ).
The expressions above for a X a" and a" yield
(a X a"). a" = 6182.
It remains only to substitute this data into the formulas in Theorem 4.3,
with N being computed by another cross product. The final results are
T =
N =
B =
(1  t\ 2/, 1 + p)
V2(l + t 2 )
(2U  * 2 ,0)
1 + t 2
(1 + t\ 2/,l + f)
V2(l + t 2 )
1
K = T =
3(1 + Z 2 ) 2
Alternatively, we could use the identity in Lemma 1.8 to compute
 a X a" , and express
{a X a ,) •« = a • (a X a )
as a determinant by Exercise 4 of Section 1.
Sec. 4]
ARBITRARYSPEED CURVES
71
FIG. 2.17
Let us summarize the situation. We now have the Frenet apparatus for
an arbitraryspeed curve a. This apparatus satisfies the extended Frenet
formulas (with factor v) and may be computed by Theorem 4.3. If v = 1,
that is, if a is a unitspeed curve, the Frenet formulas in Lemma 4.1 simplify
slightly (to Theorem 3.2), but Theorem 4.3 may be replaced by the much
simpler definitions in Section 3.
Let us consider some applications of the results in this section. There
are a number of interesting ways in which one can assign to a given curve
/3 a new curve $ whose geometric properties illuminate some aspects of the
behavior of /3. For example, if jS is a unitspeed curve, the curve a = T
is the spherical image of /3. According to Remark 3.4, a is the curve
such that each point a(s) has the same Euclidean coordinates as the unit
tangent vector T(s) (Fig. 2.17). Roughly speaking, (r(s) is gotten by mov
ing T(s) to the origin. The spherical image lies entirely on the unit sphere
2 of E , since [<r = \\ T \\ = 1, and the motion of <x represents the curving
of P.
For example, if £ is the helix in Example 3.3, the formula there for T
shows that
/N / a . s a s b\
r{S) = I sin  ,  COS  ,  J
\ c c c c c)
Thus the spherical image of a helix lies on the circle cut from the unit
sphere by the plane z = b/c.
There is no loss of generality in assuming that the original curve /? has
unit speed, but we cannot also expect a to have unit speed. In fact, since
a = T, we have <r = T = kN. Thus a moves always in the principal
normal direction of /3, with speed  a  equal to the curvature k of /?.
Next we assume k > 0, and use the Frenet formulas for to compute the
curvature of a. Now
as
. K *T + ^ N + ktB.
as
Thus
72
FRAME FIELDS
[Chap. II
* X <;" =  K 3 N X T + ktN X B = k 2 ( k £ + rT).
By Theorem 4.3 the curvature of the spherical image a is
 II *' X a" 11 _ \//c 2 + r 2
(■+
2\l/2
> 1
and thus depends only on the ratio of torsion to curvature for the original
curve 0.
Here is a closely related application in which this ratio t/k turns out to
be decisive.
4.5 Definition A regular curve a in E 3 is a cylindrical helix provided the
unit tangent T of a has constant angle # with some fixed unit vector u; that
is, T(t)»u = cos #for alU.
This condition is not altered by reparametrization, so for theoretical
purposes we need only deal with a cylindrical helix which has unit speed.
So suppose is a unitspeed curve with T'u = cos &. If we pick a reference
point, say 0(0), on 0, then the realvalued function
h(s) = (0(s) 0(O)). U
tells how far 0(s) has "risen" in the u direction since leaving 0(0) (Fig.
2.18). But
dh
ds
= »u = T»n = cos #
so is rising at a constant rate relative to arc length, and h(s) = s cos #.
(If we shift to an arbitrary parametrization, this formula becomes
h(t) = s(t) cos #,
where s is the arclength function.)
FIG. 2.18
FIG. 2.19
Sec. 4] ARBITRARYSPEED CURVES 73
By drawing a line through each point of j8 in the u direction, we construct
a generalized cylinder C on which ft moves in such a way as to cut each
ruling (or "element") at constant angle #, as in Fig. 2.19. In the special
case when this cylinder is circular, /3 is evidently a helix of the type defined
in Example 3.3.
It turns out to be quite easy to identify cylindrical helices.
4.6 Theorem A regular curve a with k > is a cylindrical helix if and
only if the ratio t/k is constant.
Proof. It suffices to consider the case where a has unit speed. If a is a
cylindrical helix with T*n = cos &, then
= (Tu)' = T'*u = kNu.
Since k > 0, we conclude that N*u = 0. Thus for each s, u lies in the
plane determined by T(s) and B(s). Orthonormal expansion yields
u = cos # T + sin # B.
As usual we differentiate and apply Frenet formulas to obtain
= (k cos & — t sin &)N.
Hence t sin # = k cos t?, so that t/k has constant value cot t?.
Conversely, suppose that t/k is constant. Choose an angle # such that
cot # = t/k. If
U = cos # T + sin # £,
we find
£/' = (k cos i?  t sin d)N = 0.
This parallel vector field U then determines (as in Remark 3.4) a unit
vector u such that T*\i = cos t?, so a is a cylindrical helix. 
This proof also shows how to compute the unit vector u and angle t?.
For example, the curve a in Example 4.4 is a cylindrical helix, since k = r.
The angle # satisfies the equation cot # = t/k = 1 ; we take & = 7r/4.
Then cos # = sin # = 1/V2, so by the proof above, u = (1/V2) (7 1 + #)•
The data in Example 4.4 then yield u = (0,0, 1). (There is no need to
convert a to unit speed — that would merely reparametrize k, t, T, and B,
with no effect on t? and u. )
In Exercise 10 this information about cylindrical helices is used to show
that circular helices are characterized by constancy of curvature and tor
sion (see also Corollary 5.5, of Chapter III).
Simple hypotheses on a regular curve in E 3 thus have the following
effects (<=> means "if and only if")
74 FRAME FIELDS [Chap. II
k = <=> straight line
r = <=> planar
k constant > and t = «=> circle
k constant > and r constant ?^ <=> circular helix
t/k constant <=> cylindrical helix
EXERCISES
1. Consider the curve a: R — ► E 3 such that a(J) = (2t, t 2 , f/3).
(a) Compute the Frenet apparatus of a: k, t, T, N, B.
(b) Make a careful sketch of this curve for — 4 ^ t ^ 4 showing
7\ JV, and 5 at t = 0, 2, 4. (#tnfc Begin with its projection (2t, t\ 0)
in the xy plane.)
(c) Find the limiting position of the Frenet frame T, N, B of a as
t —> + oo and 2 — > — oo .
2. Compute the Frenet apparatus of the curve a(t) = (cosh t, sinh t, t).
Express the curvature and torsion of a as functions k(s) and t(s)
of arc length s measured from t = 0.
3. For the curve a(t) = (t cos t, tsint,t),
(a) compute the Frenet apparatus at t = 0. (Evaluate a, a , ol" at
t = before using Theorem 4.3.)
(b) sketch this curve for — 2ir ^ t ^ 2*, showing T, N, B a,t t = 0.
(Hint: Ex. 2 of II.2.)
4. For the curve a in Example 4.4, check Lemma 4.2 by direct substi
tution. Make a sketch, in scale, showing the vectors T(0), N(0), a (0),
anda"(0).
5. Prove that the curvature of a regular curve in E 3 is given by
kV =  a"  2  (dv/dtf.
6. If a is a curve with constant speed c > 0, show that
T = a'/c k =  a" ll/c 2
N = a"/\\ a"  _ a X a" 'a'"
B = «' X a"A  a"  T ~ c 2  a"  2
where for N, B, r, we assume a" never zero, that is, k > 0.
7. Use the formulas in the preceding exercise to compute the Frenet
apparatus of the helix a in Example 4.2 in Chapter I.
8. Let a be a cylindrical helix with unit vector u, angle #, and arclength
function s (measured from, say, t = 0. ) The unique curve 7 such that
Sec. 4] ARBITRARYSPEED CURVES 75
«(0 = 7(0 + s(t) cos #u is called the crosssection curve of the cylin
der on which a lies. Prove that
(a) 7 lies in the plane through a(0) orthogonal to u.
(b) The curvature of 7 is /c/sin 2 d, where a is the curvature of a.
(Hint: For (b) it suffices to assume a has unit speed.)
9. (Continuation). The following curves are cylindrical helices; for each
find the unit vector u, angle #, and crosssection curve 7; verify con
dition (a) above.
(a) The curve in Exercise 1.
(b) The curve in Example 4.4.
(c) The curve in Exercise 2.
70. If is a unitspeed curve with k > and r 5* both constant, prove
that jS is a (circular) helix.
11. Let a be the spherical image (Section 4) of a unitspeed curve /?.
Prove that the curvature and torsion of a are
^ / 1 _l ( / \s (d/ ds)(r/K)
41 + W#cn
where « and r are the curvature and torsion of /3.
12. (a) Prove that a curve is a cylindrical helix if and only if its spherical
image is part of a circle. (No computations needed.)
(b) Sketch the spherical image of the cylindrical helix in Exercise 1.
Is it a complete circle? Find its center.
13. If a is a curve with k > 0, then the central curve a* = a + (\/k)N
consists of all centers of curvature of a (Ex. 6 of II.3.) For any two
nonzero numbers a and b, let @ a b be the helix in Example 3.3. Show that
the central curve of /8a*. is Pab, where a = — b 2 /a. Deduce that the cen
tral curve of && is the original helix /3„ 6 .
14. If a(t) = (x(t), y(t)) is a regular curve in E 2 , show that its curvature
(Ex. 8, II.3) is
= a".J(d) = xy"  x"y
K v* (x' 2 + y' 2 ) 3 ' 2 '
Here J is the operator such that
J(tiM) = (Mi)
15. For a regular curve a in E 2 , the central curve a* = a + (l/*c)iV"is
called the evolute of a. (It is, of course, not defined at points where
K = 0.)
(a) Show that a* is uniquely determined by the condition that its
tangent line at each point a*(t) is the normal line to a at a(t).
76 FRAME FIELDS [Chap. II
(b) Prove that
f i
a »J (a )
(notation as in Exercise 14)
(c) Find the evolute of the cycloid
«(0 = (t + sin tyl + cos 0, — t < t < t.
Sketch both curves.
16. The total curvature of a unitspeed curve a denned on I, is ji k(s) ds
(an improper integral when the interval is infinite). If a is merely
regular, the definition becomes f t n(t)v(t) dt; this makes total curva
ture independent of the parametrization of a. Find the total curvature
of the following curves — the first three defined on the whole real line.
(a) The curve in Example 4.4.
(b) The helix in Example 3.3.
(c) The curve in Exercise 2.
(d) The ellipse a{t) = (a cos t, b sin t). Since this is a closed curve,
consider only a single period ^ t £ 2t.
17. Let / > and g be arbitrary differentiable realvalued functions on
some interval in R. Consider the curve
«(0 = (//(*> sin *.//«) cos t,ff(t)g(t))
where jh denotes any function whose derivative is h. Show that the
curvature and torsion of a are given by
W 1 *
+ <7 2 + <7' 2 1 q + g"
T =
+ <7 2 ) 3 / (i + g 2 + g' 2 )'
18. Consider the general cubic curve y(t) = (at, bt 2 , ct 3 ), where abc ^ 0.
(a) Compute
/ = ?¥ f 9c¥ + 4bV + q 2 T /2
T/K 26 2 _9c 2 * 4 + 9(a 2 c 2 /6 2 )* 2 + a 2 J
and deduce that the cubic curve 7 is a cylindrical helix if and only
if Sac = ± 26 2 .
(b) In the case where Sac = 2b 2 , find the unit vector u and angle #.
19. One of the standard tricks of advanced calculus is the construction of
an (infinitely) differentiable function / on the real line such that
f(t) = for t ^ 0, and f(t) > for t > 0. (Also.f(0 > for t > 0.)
If g(t) = f( — t), consider the curve
a(t) = (t,f(t),g(t)).
Sec. 5}
COVARIANT DERIVATIVES
77
(a) Prove that the curvature of a is zero only when t = 0.
(b) Sketch this curve for  t  small, and show some principal normals
for t > and t < 0.
This example shows that the condition k > cannot be avoided in a
detailed study of the geometry of curves in E 3 , for if k is zero even at a
single point, the geometric character of the curve can change radically at
that point. (Note that this difficulty is not a serious one for curves in E 2 ;
see Ex. 8 of II.3.)
5 Cova riant Derivatives
In Chapter I, each time we defined a new object (curve, differential form,
mapping, . . .) we usually followed by defining an appropriate notion of
derivative of the object. Vector fields were an exception; we have delayed
defining their derivatives since (as later results will show) this notion
belongs properly to the geometry of Euclidean space.
The definition generalizes that of the derivative v[/] of a function /
with respect to a tangent vector v at a point p (Definition 3.1 of Chapter
I). In fact, replacing / by a vector field W, we observe that the function
t — > W(p + tv) is a vector field on the curve t — ► p + tv. (The derivative
of such a vector field was defined in Section 2.) Now the derivative of W
with respect to v will be the derivative of t — > W(p + tx) at t = 0.
5.1 Definition Let W be a vector field on E 3 and let v be a tangent vector
to E 3 at the point p. Then the covariant derivative of W with respect to v
WU)
^ V W
FIG. 2.20
78 FRAME FIELDS [Chap. II
is the tangent vector
V V W = W(p + tv)'(0)
at the point p.
Evidently V V W measures the initial rate of change of W(p) as p moves
in the v direction (Fig. 2.20). (The term "covariant" derives from the
generalization of this notion discussed in Chapter VII.)
For example, suppose W = xU x + yzU%, and v = ( — 1,0, 2) at
P= (2,1,0).
Then
p + tv = (2  t, 1, 20,
so
W(p + tv) = (2  t) 2 U x + 2tU 3 ,
where strictly speaking Ui and t/ 2 are also evaluated at p + tv. Thus,
V V W = W( V + «v)'(0) = 4C/ 1 (p) + 2C^,(p).
5.2 Lemma If W = E «>,•£/,• is a vector field onE 3 , and v is a tangent
vector at p, then
V V W = £vK] Ui(p).
Proof. We have
W(p + tx) = X) «i(p + <v) tf.(p + tv)
for the restriction of W to the curve t — > p + tv. To differentiate such a
vector field (at £ = 0), one simply differentiates its Euclidean coordinates
(at t = 0). But by the definition of directional derivative (Definition 3.1
of Chapter I), the derivative of w\(p + tv) at t = is precisely v[w t ].
Thus
V V W = W(p + «v)'(0) = £ vN Ui(p). 
In short, to apply V„ to a vector field, apply v to its Euclidean coordinates.
Thus the following linearity and Leibnizian properties of covariant deriva
tive follow easily from the corresponding properties (Theorem 3.3 of Chap
ter I ) of directional derivatives.
5.3 Theorem Let v and w be tangent vectors to E 3 at p, and let Y and
Z be vector fields on E 3 . Then
(1) Vav+bwY — aV v Y + bV w Y for all numbers a and b.
(2) V v (aY + bZ) = aV„Y + bV v Z for all numbers a and b.
Sec. 5] COVARIANT DERIVATIVES 79
(3) V„(/F) = v[/]F(p) +/(p) V„Fforall (differentiable) functions/.
(4) v[YZ] = V„7.Z(p) + F(p).V,Z.
Proof. For example, let us prove (4). If
Y = 22 ViUi and Z = ^T, ZiUi,
then
Hence by Theorem 3.3 of Chapter I,
v[FZ] = v[£ y«*] = £ vfo] z t (p) + £ !fc(p) vW
But by the preceding lemma,
V r F = ZvIi/JC/^p) and V r Z = 2>NWp)
Thus the two sums displayed above are precisely V„F»Z(p) and F(p) • V„Z.
Using the pointwise principle (Chapter I, Section 2) once again, we can
take the covariant derivative of a vector field W with respect to a vector
field V, rather than a single tangent vector v. The result is the vector field
V V W whose value at each point p is V r ( P )W. Thus V r W consists of all the
covariant derivatives of W with respect to the vectors of V. It follows
immediately from the lemma above that if W = ^ Willi, then
V Y W = £ V[ Wi ]Ui.
Coordinate computations are easy using the basic identity £/*[ /] = df/dXi.
For example, suppose V = (y — x)U x + xyU 3 and (as in the example
above) W = xU x + yzU z . Then
V[x 2 ] = (y  x)UAx] = 2x(y  x)
V[yz] = xyU 3 [yz] = xy 2 .
Hence
V V W = 2x(y  x)U! + xy 2 U z .
Now the vector field V was also selected with the earlier example in mind.
In fact, the value of V at p = (2, 1, 0) is
V(p) = (1  2)C/ 1 (p) + 2*7 3 (p) = (1,0,2), = v p ,
as before. Thus the value of the vector field V V W at this point p must
agree with the earlier computation of V V W. And, in fact, for p = (2,1,0),
Vr(W0(p) = 22(1  2)C/ 1 (p) + 2£/,(p) = Wi(p) + 2U t (p).
80 FRAME FIELDS [Chap. II
For the covariant derivative V r W as expressed entirely in terms of vector
fields, the properties in the preceding theorem take the following form.
5.4 Corollary Let V, W, F, and Z be vector fields on E 3 . Then
(1) W(aF + bZ) = aV Y Y + bV v Z, for all numbers a and b.
(2) V f v+gwY = fV v Y + gV w Y, for all functions / and g.
(3) V v (fY) = V[f]Y + fV v Y, for all functions/.
(4) V[YZ] = VyYZ + FVvZ.
We shall omit the proof, which is an exercise in the use of parentheses
based on the (pointwise principle) definition (V F F)(p) = vV (p )F.
Note that V F F does not behave symmetrically with respect to V and F.
This is to be expected, since it is F that is being differentiated, while the
role of V is merely algebraic. In particular, V /r F is/V r F, but V y (/F) is not
/V r F: There is an extra term arising from the differentiation of / by V.
EXERCISES
1. Consider the tangent vector v = (1,1,2) at the point p = (1,3, — 1).
Compute V V W directly from the definition, where
(a) W = x 2 U x + yU*. (b) W = xU x + x 2 U 2  z 2 U z .
2. Let V = — ylli + xU 3 and W = cos xUi + sin xXJi.
Express the following covariant derivatives in terms of Ui, U%, Us:
(a) V Y W. (c) V Y {zW). (e) V v (V r W).
(b) V r V. (d) V W (V). (f) V v (xV  zW).
3. If W is a vector field with constant length  W , prove that for any
vector field V, the covariant derivative V V W is everywhere orthogonal
to IF.
4. Let X be the special vector field ^3 %iUi, where x\, x 2 , x 3 are the natural
coordinate functions of E 3 . Prove that V r X = V for every vector field
V.
5. If W = 52 Willi is a vector field on E 3 , the covariant differential of W is
defined to be VW = ^ dwi Ui. Here VW is the function on all tangent
vectors whose value on v is
J2dWi(v)Ui(p) = V V W.
Compute the covariant differential of
W = :
and use it to find V„W, where
W = xy Ui — x 2 z 2 Uz,
Sec. 6] FRAME FIELDS 81
(a) v= (1,0, 3) at p= (1,2,1).
(b) v= (1,2,1) at p = (1,3,2).
6. Let W be a vector field defined on a region containing a curve a. Then
t — > W(a(t)) is a vector field on a called the restriction of W to a and
denoted by PF«.
(a) Prove that V a > it) W = (JF«)'(0
(b) Deduce that the straight line in Definition 5.1 may be replaced by
any curve with initial velocity v. Thus the derivative Y of a vector
field Y on a curve a is (almost) V a 'F.
7. The bracket of two vector fields is the vector field [V, W] = V F W — V W V.
Establish the following properties of the bracket:
(a) [V, W][f] = VW[f]  WV[f] (here VW[f] denotes the "second
derivative" 7[TF[/]]).
(b) [W, V] = [V, W\.
(c) [U, [V, W}} + [V, [W, U]] + [W, [U, V]] = 0.
(d) [fV, gW] = fV[g]W  gW[f]V + fg[V, W).
{Hint: Z[f] = for all /implies Z = 0.)
6 Frame Fields
When the Frenet formulas were discovered (by Frenet in 1847, and inde
pendently by Serret in 1851), the theory of surfaces in E 3 was already a
richly developed branch of geometry. The success of the Frenet approach
to curves led Darboux (around 1880) to adapt this "method of moving
frames" to the study of surfaces. Then, as we mentioned earlier, it was
Cartan who brought the method to full generality. His essential idea was
very simple: To each point of the object under study (a curve, a surface,
Euclidean space itself, . . .) assign a frame; then using orthonormal ex
pansion express the rate of change of the frame in terms of the frame itself.
This, of course, is just what the Frenet formulas do in the case of a curve.
In the next three sections we shall carry out this scheme in detail for
the Euclidean space E 3 . We shall see that geometry of curves and surfaces
in E is not merely an analogue, but actually a corollary, of these basic
results. Since the main application (to surface theory) comes only in
Chapter VI, these sections may be postponed, and read as a preliminary
to that chapter.
By means of the pointwise principle (Chapter I, Section 2) we can
automatically extend operations on individual tangent vectors to opera
tions on vector fields. For example, if V and W are vector fields on E 3 , then
82
FRAME FIELDS
[Chap. II
the dot product V'W of V and W is the (differentiable) realvalued
function on E whose value at each point p is F(p)» W(p). The norm \\V\\
of V is the real valued function on E 3 whose value at p is  F(p). Thus
 F  = (VV) m . By contrast with V*W, the norm function  V  need
not be differentiable at points for which V(p) = 0, since the squareroot
function is badly behaved at 0.
At each point p of E 3 , the three tangent vectors J7i(p), U 2 (p), U 3 (p)
constitute a frame at p. This remark is concisely expressed in terms of dot
products of vector fields by writing UrUj = 5,y (1 ^ i,j ^3). We used
U h U 2 , U 3 throughout Chapter I. Now, with the dot product at our dis
posal, we can introduce a simple but crucial generalization.
6.1 Definition Vector fields E t , E 2 , E 3 on E 3 constitute a frame field on
E provided
Ei*Ej = Si
where 8^ is the Kronecker delta.
(1 ^ i,j ^ 3)
The term frame field is justified by the fact that at each point p the
three vectors i?i(p), E 2 (p), E 3 (p) form a frame at p. We anticipated this
in Chapter I by calling U\, 11%, U s the natural frame field on E 3 .
6.2 Example (1) The cylindrical frame field (Fig. 2.21). Let r, &, z be
the usual cylindrical coordinate functions on E 3 . We shall pick a unit vector
field in the direction in which each coordinate increases (when the other
two are held constant). For r, this is evidently
Ei = cos d Ui + sin # U 2 ,
pointing straight out from the z axis. Then
E 2 =  sin # Ui + cos # U 2
points in the direction of increasing & as in Fig. 2.21. Finally the direction
Cylindrical
FIG. 2.21
Spherical
FIG. 2.22
Sec. 6]
FRAME FIELDS
83
E 3 = U 3
FIG. 2.23
of increase of z is, of course, straight up, so
E 3 = U 3 .
It is easy to check that Ei'Ej = 8a, so this is a frame field (defined on
all of E except the z axis). We call it the cylindrical frame field on E 3 .
(2) The spherical frame field on E 3 (Fig. 2.22). In a similar way, a frame
field Fi, F 2 , F 3 can be derived from the spherical coordinate functions
p, #, <p on E . As indicated in the figure, we shall measure <p up from the
xy plane rather than (as is usually done) down from the z axis.
Let Ei, E 2 , E 3 be the cylindrical frame field. For spherical coordinates, the
unit vector field F 2 in the direction of increasing
# is the same as above, so F 2 = E 2 . The unit vec
tor field JPi, in the direction of increasing p, points
straight out from the origin; hence it may be ex
pressed as
^1 = cos <p Ei f sin <p E 3 ,
(Fig. 2.23). Similarly the vector field for increas
ing <p is
^3 = —sin <p Ei + cos <p E 3 .
Thus the formulas for E u E 2 , E 3 in (1 ) yield
Fi = cos <p(cos # Ui + sin # U 2 ) + sin <p XJ%
F 2 = sin # C7i + cos d U 2
F 3 = —sin <p(cos & Ui + sin $ U 2 ) + cos <p U 3 .
By repeated use of the identity sin 2 + cos 2 = 1, we check that F h F 2 , F 3
is a frame field — the spherical frame field on E 3 . (Its actual domain of defi
nition is E minus the z axis, as in the cylindrical case.)
The following useful result is an immediate consequence of orthonormal
expansion.
6.3 Lemma Let E h E 2 , E 3 be a frame field on E 3 .
(1 ) If V is a vector field on E 3 , then V = X) /*^« where the functions
ft — V'Ei are called the coordinate functions of V with respect to E i} E 2 , E z .
(2) If V = YtfiEi and W = E 9<Ei, then VW = X)/^. In par
ticular,  v  = (Z//) 1/2 
Thus a given vector field V has a different set of coordinate functions
with respect to each choice of a frame field E u E 2 , E 3 . The Euclidean co
ordinate functions (Lemma 2.5 of Chapter I), of course, come from the
natural frame field U u U 2 , U 3 . In Chapter I, we used this natural frame
field exclusively, but now we shall gradually shift to arbitrary frame fields.
84
FRAME FIELDS
[Chap. II
The reason is clear: In studying curves and surfaces in E 3 , we shall then be
able to choose a frame field specifically adapted to the problem at hand. Not
only does this simplify computations, but it gives a clearer understanding
of geometry than if we had insisted on using the same frame field in every
situation.
EXERCISES
1. If V and W are vector fields on E 3 which are linearly independent at
each point, show that
Ei =
V
V
E» =
W
w\\'
Ez = Ei X E%
is a frame field, where W = W — W'EiEi.
2. Express each of the following vector fields (i) in terms of the cylindrical
frame field (with coefficients in terms of r, #, z) and (ii) in terms of the
spherical frame field (with coefficients in terms of p, #, <p) :
(a) Ui. (b) cos # Ui + sin # U 2 + U».
(c) xUt + yU* + zUz.
3. Find a frame field E\, E 2 , E 3 such that
E l = cos x Ui + sin x cos z[/ 2 + sin x sin z £/ 3 .
4. Toroidal frame field. Let be all of E 3 except the z axis and the circle C
of radius R in the xy plane. The toroidal coordinate functions p, #, <p
are defined on as suggested in Fig. 2.24, so that
FIG. 2.24
Sec. 7] CONNECTION FORMS 85
x = (R + p cos ^>) cos t?
?/ = (R + p cos p) sin #
z = p sin <e.
If Ei, E 2 , and E 3 are unit vector fields in the direction of increasing
P, #, and <p, respectively, express E h E 2 , E 3 in terms of U h U 2 , U 3 and prove
that it is a frame field.
7 Connection Forms
Once more we state the essential point: The power of the Frenet formulas
stems not from the fact that they tell what the derivatives T' , N', B f are,
but from the fact that they express these derivatives in terms of T, N, B —
and thereby define curvature and torsion. We shall now do the same thing
with an arbitrary frame field E u E 2 , E 3 on E 3 ; namely, express the covariant
derivatives of these vector fields in terms of the vector fields themselves. We begin
with the covariant derivative with respect to an arbitrary tangent vector
v at the point p. Then
V v Ei = CnEiip) + c 12 E 2 (p) + Ci3# 3 (p)
V V E 2 = CuEiip) + C22# 2 (p) + C23# 3 (p)
V V E 3 = C 3 l#l(p) + C32# 2 (p) + C33^ 3 (p)
where by orthonormal expansion the coefficients of these equations are
dj = V v Ei*Ej(p) for 1 S i,j ^ 3.
These coefficients c,y depend on the particular tangent vector v, so a
better notation for them is
co ti (v) = VvEiEjip), (1 ^ i,j ^ 3).
Thus for each choice of i and j, on, is a real valued function defined on all
tangent vectors. But we have met that kind of function before.
7.1 Lemma Let E u E 2 , E 3 be a frame field on E 3 . For each tangent vector
v to E 3 at the point p, let
« tV (v) = VvEiEjip), (1 ^ i,j ^ 3).
Then each «,,■ is a 1form, and co t7 = — Wjl . These 1 forms are called
the connection forms of the frame field E u E 2 , E 3 .
Proof. By definition, &>,,■ is a realvalued function on tangent vectors,
so to verify that w.y is a 1form (Definition 5.1 of Chapter I), it suffices to
86 FRAME FIELDS [Chap. II
check the linearity condition. But using Theorem 5.3, we get
03ij(a\ + bw) = V av+bw Ei'Ej(p)
= (aV v Ei + bVvE^Ejip)
= aV.Ei'Ej(p) + bVvErEjtp)
= aa)ij(x) + bcoij(w).
To prove that co»y = — a>y» we must show that w,y(v) = — ajyi(v) for
every tangent vector v. By definition of frame field, Ei*Ej = 8^, and since
each Kronecker delta has constant value or 1, we have \[8i 3 ] = 0. Thus
by the Leibnizian formula (4) of Theorem 5.3,
= vWfEj] = VyEiEjip) + Ei(p).V r Ej.
By the symmetry of the dot product, the two vectors in this last term may
be reversed, so we have found that = &>»y(v) + w 7t (v). 
The geometric significance of the connection forms is no mystery. The
definition «,v(v) = V Y Ei*Ej(p) shows that co,v(v) is the initial rate at
which E{ rotates toward Ej as p moves in the v direction. Thus the 1forms
Ma contain this information for all tangent vectors to E !
The following basic result is little more than a rephrasing of the defini
tion of connection forms.
7.2 Theorem Let co;j (1 ^ i,j ^ 3) be the connection forms of a frame
field E x , E 2 , E z on E 3 . Then for any vector field V on E 3 ,
VvEi = E <*ii(y)E h (1 S i ^ 3).
j
We call these the connection equations of the frame field E x , E 2 , E z .
Proof. For fixed i, both sides of this equation are vector fields. Thus we
must show that at each point p,
Vv,p)^ = E ««(F(p))^y(p).
But as we have already seen, the very definition of connection form makes
this equation a consequence of orthonormal expansion. 
When i = j, the skewsymmetry condition <«>„ = — wyi becomes
thus
COn = C022 == W33 = 0.
Hence this condition has the effect of reducing the nine 1forms Wjy for
Sec. 7] CONNECTION FORMS 87
1 ^ i,j ^ 3 to essentially only three, say co^, wu, W23. It is perhaps best to
regard the connection forms «,,■ as the entries of a skew symmetric matrix
of 1 forms,
(COn «12 Wu\ J C012 C0l3\
0>21 W22 ^23 I = I _ Wl2 0>23
W31 W S 2 OJ33/ \ — 0)13 — U)23
Thus in expanded form, the connection equations (Theorem 7.2) become
V v Ei = coi2(V)Ei f &>i3(V)2?3
V y # 2 = C0 12 (F)^! + tt»(7).Bj
Vy^J 3 = — (t)n(V)Ei — 0>23(V)^2.
showing an obvious relation to the Frenet formulas
T' = kN
N' = kT + tB
B' =  rN.
The absence from the Frenet formulas of terms corresponding to un{V)E 3
and — wu (V) Ei is a consequence of the special way the Frenet frame field
is fitted to its curve. Having gotten T(~Ei), we chose N(~E 2 ) so that
the derivative T' would be a scalar multiple of N alone and not involve
B(~E t ).
Another difference between the Frenet formulas and the equations above
stems from the fact that E 3 has three dimensions, while a curve has but one.
The coefficients — curvature k and torsion t — in the Frenet formulas measure
the rate of change of the frame field T, N, B only along its curve, that is,
in the direction of T alone. But the coefficients in the connection equations
must be able to make this measurement for E ly E i} E% with respect to ar
bitrary vector fields in E 3 . This is why the connection forms are 1 forms and
not just functions.
These formal differences aside, a more fundamental distinction stands
out. It is because a Frenet frame field is specially fitted to its curve that
the Frenet formulas give information about that curve. Since the frame
field Ei, E 2 , E 3 used above is completely arbitrary, the connection equations
give no direct information about E 3 , but only information about the "rate
of rotation" of that particular frame field. This is not a weakness, but a
strength, since as indicated earlier, if we can fit a frame field to a geometric
problem arising in E 3 , then the connection equations will give direct in
formation about that problem. Thus these equations play a fundamental
role in all the differential geometry of E 3 . In particular, the Frenet formulas
88 FRAME FIELDS [Chap. II
can be deduced from them (Exercise 8). For the sake of motivation, however,
we have preferred to deal with the simpler Frenet case first.
Given an arbitrary frame field E x , E 2 , E 3 on E 3 , it is fairly easy to find an
explicit formula for its connection forms. First use orthonormal expansion
to express the vector fields Ei, E 2 , E 3 in terms of the natural frame field
Ur, U 2 , C/ 3 onE 3 :
Ei = a n Ui + a 12 U 2 + a n U 3
E 2 = (hiUi + ChiUi + ckaUz
Ez = CL31U1 + a®,U 2 + azsUz.
Here each a t y = Ei* U, is a realvalued function on E 3 . The matrix
(«n «i2 aiz^
«2i an a 2 z
azi a 3 2 033/
with these functions as entries is called the attitude matrix of the frame
field Ei, E 2 , E s . In fact, at each point p, the numerical matrix
A(p) = (o,,(p))
is exactly the attitude matrix, of the frame 2?i(p), E 2 (p), E 3 (p) as in Defi
nition 1.6. Since attitude matrices are orthogonal, the transpose l A of A
is equal to its inverse A 1 .
Define the differential of A = {an) to be dA = (da t y), so dA is a matrix
whose entries are 1 forms. We can now give a simple expression for the
connection forms in terms of the attitude matrix.
7.3 Theorem If A = (a t y) is the attitude matrix and w = (co t y) the
matrix of connection forms of a frame field E u E 2 , E z , then
03 = dA l A (matrix multiplication),
or, equivalently
«»y = 2 a ik da ik for 1 ^ i,j ^ 3.
k
Proof. If v is a tangent vector at p, then by definition,
« t7 (v) = VvEiEjip).
Since A is the attitude matrix,
Ei = X) <**U k ,
and thus, by Lemma 5.2,
Sec. 7] CONNECTION FORMS 89
VvEi = £v[o ft ] U k (p).
The dot product of this vector with
Ej(p) = 2> ifc (p) U k (p)
is then
k
But by the definition of differential,
v[a lfc ] = da ik {\);
hence
w »i(v) = ^2a jk (p) da ik (\) = (^ a jfc da tfc ) (v).
Since this equation holds for all tangent vectors, the two 1 forms w»_, and
53 o,jk da ik are equal. It is easy to get the neater matrix formula. In fact,
the transpose l A has entries l a k j = aj k , so
«»•; = 2 da ik l a k j for 1 ^ ij ^ 3,
which in terms of matrix multiplication is just co = dA l A. 
Using this result, let us compute the connection forms of the cylindrical
frame field in Example 6.2. From the definition, we read off the attitude
matrix
cos & sin # 0^
A =  — sin# cost?
L
Thus
Since
'sintfdtf costfdtf 0^
dA =  — costfdtf — sintfdtf
Oy
''cost? — sint? 0^
*A = ( sin tf cos # ) ,
1,
90 FRAME FIELDS [Chap. II
we easily compute
d& 0^
a = dA l A =  d&
0/
Thus o>i2 = d& and all the other connection forms (except, of course,
co 2 i = — co i2 ) are zero. Then the connection equations (Theorem 7.2) of the
cylindrical frame field become
VyEi = d&(V)E 2 = V[#]E 2
V V E 2 = d&(V)E l = V[&\Ei
V V E 3 =
for all vector fields V.
These equations have obvious geometric significance. The third equa
tion says that the vector field E 3 is parallel. We knew this already, since in
the cylindrical frame field, E 3 is just U 3 . The first two equations say that
the covariant derivatives of E x and E% with respect to an arbitrary vector
field V depend only on the rate of change of the angle & in the direction of
V. From the way the function # is defined, it is clear that V[&] = when
ever V is, at each point, tangent to the plane through the z axis. Thus for
a vector field of this type, the connection equations above predict that
V v Ei = V v Ei = 0. From Fig. 2.21, it is clear that E 1 and E 2 do stay parallel
on any plane through the z axis.
EXERCISES
1. For any function /, show that the vector fields
E, = (sin fU, + U*  cosfU 3 )/V2
E 2 = (sin/C/x  U*  cosfU 3 )/V2
E z = cos fU\\ sin / U 3
form a frame field, and find its connection forms.
2. Find the connection forms of the natural frame field Ui, U 2 , U 3 .
3. For any function /, show that
cos /
cos / sin /
sin/
sin / cos /
• 2 e
sm /
cos/
sin/
cos/
is the attitude matrix of a frame field, and compute its connection
forms.
Sec. 8] THE STRUCTURAL EQUATIONS 91
4. Prove that the connection forms of the spherical frame field are
wi2 = cos <p d&, coi3 = d<p, C023 = sin <p d&.
5. If Ei, E 2 , E s is a frame field and W = ^ fiE i? prove the covariant de
rivative formula:
v r w = Zmfj] + I/wTO.
J i
6. Let Ei, Ei, E z be the cylindrical frame field. If V is a vector field such
that V[&] = 1, compute vV(r cos # #1 + r sin # ^/ 3 ).
7. If Fi, F 2 , Fz is the spherical frame field,
(a) Prove that Fi[p] = 1 and FM] = FM = 0.
(b) Compute V, x (cos p F 2 + sin p F 3 ).
8. Let j8 be a unitspeed curve in E 3 with k > 0, and suppose that i?i, E 2 , E 3
is a frame field on E 3 such that the restriction of these vector fields to
8 gives the Frenetframe field T, N, B of 0. Prove that
Mir) = k, W i,(r) = o, co 23 (r) = r.
Then deduce the Frenet formulas from the connection equations. {Hint:
Ex. 6 of II.5.)
8 The Structural Equations
We have seen that 1 forms — the connection forms — give the simplest
description of the rate of rotation of a frame field. Furthermore, the frame
field itself can be described in terms of 1 forms.
8.1 Definition If E x , E 2 , E 3 is a frame field on E 3 , then the dual lforms
0i, 02, 03 of the frame field are the lforms such that
Bi(y) = vEi(p)
for each tangent vector v to E at p.
Note that 0» is linear on the tangent vectors at each point; hence it is a
1form. (Readers familiar with the notion of dual vector spaces will recog
nize that at each point, 0i, 2 , 03 gives the dual basis of E u E 2 , E 3 .)
In the case of the natural frame field Ui, U 2 , Uz, the dual forms are just
dxi, dx 2 , dx 3 . In fact, from Example 5.3 of Chapter I we get
dxi{\) = Vi = v[/,(p)
for each tangent vector v; hence dxi = t .
Using dual forms, the orthonormal expansion formula in Lemma 6.3
92 FRAME FIELDS [Chap. II
may be written V = E 9i{V)Ei. In the characteristic fashion of duality,
this formula becomes the following lemma.
8.2 Lemma Let 0i, 2 , 03 be the dual 1 forms of a frame field E u E 2 , E 3 .
Then any 1form <j> on E has a unique expression
Proof. Two 1 forms are the same if they have the same value on any
vector field V. But
(2>w*i)(J0 = 2>CE<)fc<n
These functions <f>(Ei) are the only possible coordinate functions for <j> in
terms of 6 U B 2 , 03, since if <f> = E fid iy then
<f>(Ej) = E fMEj) = E /Ay = /y I
Thus <f> is expressed in terms of dual forms of Ei, E 2 , E 3 by evaluating it
on Ei, E2, E 3 . This useful fact is the generalization to arbitrary frame fields
of Lemma 5.4 of Chapter I.
We compared a frame field E u E 2 , E 3 to the natural frame field by means
of its attitude matrix A = {an), for which
Ei = E *aUi (l£i£ 3).
The dual formulation is just
6i — E °*'i dxj
with the same coefficients. In fact, by Lemma 8.2 (or rather its special case,
Lemma 5.4 of Chapter I), we have
Oi = E 9i{Uj) dxj.
But
di(Uj) = EiUj = (E aikU k )'Uj = an.
These formulas for Ei and 0; show plainly that 0i, 2 , 3 is merely the dual
description of the frame field E u E 2 , E 3 .
In calculus, when a new function appears on the scene, it is natural to
ask what its derivative is. Similarly with 1forms — having associated with
each frame field its dual forms and connection forms — it is reasonable to
ask what their exterior derivatives are. The answer is given by two neat
sets of equations discovered by Cartan.
8.3 Theorem (Cartan structural equations) Let E x , E 2 , E 3 be a frame
field on E 3 with dual forms X , 2 , 03 and connection forms 0,,, (1 ^ i,j ^ 3).
The exterior derivatives of these forms satisfy
Sec. 8] THE STRUCTURAL EQUATIONS 93
(1 ) the first structural equations:
ddi = 2 «« A e > (1 ^ * ^ 3);
(2) the second structural equations:
dun = 22 «»* A w *y (1 = *J = 3).
it
Because 0, is the dual of E i7 the first structural equations may be easily
recognized as the dual of the connection equations. Only on the basis of
later experience will we discover that the second structural equation shows
that E 3 is flat — roughly speaking, in the same sense that the plane E 2 is flat.
Proof. We have seen that
hence
ddi = 22 dttij a dxj.
Since the attitude matrix A = (an) is orthogonal, the expression in Theo
rem 7.3 for wn in terms of dan may be solved for dan by the usual for
malism of linear algebra to give
dan = 2j w utQ>kj'
k
Thus
ddi = 22{ (22 uika k j) A dxj]
i k
= 22 {vat A /2 aki dxj\
k i
— 22 w * A ^fc
k
which is the first structural equation.
We could give a similar index proof for the second structural equation,
but a liberal use of matrix notation will give a better idea of what is really
going on. To apply the exterior derivative d to a matrix of functions or
1 forms, we apply it to each entry. The matrix formula go = dA l A in
Theorem 7.3, for example, means
w,y = 2^ daikajk.
But
dwij = — /] danc A dajk.
(Note the minus sign!) Hence in matrix notation, where we shall suppress
94 FRAME FIELDS [Chap. II
the wedge,
doo = dA l (dA).
Multiplying both sides of co = dA l A by A yields
dA =uA,
since A = A~\ Then the rule for transpose of a matrix product gives
l (dA) = *M) = l A l oo.
By substituting in the equation above for do: we find
do: = — coA A co = —
co co = coco
since co is skew symmetric. But this is just the second structural equation
dun = ^2 ooik A oikj in matrix notation. 
8.4 Example Structural equations for the spherical frame field (Ex
ample 6.2). The dual forms and connection forms are
0i = dp coi2 = cos <p d&
2 = p cos (p d& coi3 = dtp
3 = p dip co 2 3 = sin <p d&.
Let us check, say, the first structural equation
d8 3 = 2_, &3j A 8j = CO31 A Q 1 \ tojg a # 2 .
Using the skew symmetry co t y = — co.,; and the general properties of forms
developed in Chapter I, we get
C031 A 81 = —d<p a dp = dp a d<p
CO32 a 2 = ( — sin cp d#) a (pcos<pd#) =
(the latter since ci# a d# = 0). The sum of these terms is, correctly,
d8z = d(p d(p) = dp a dip.
Second structural equations involve only one wedge product. For ex
ample, since con = co 2 2 = 0,
dun = 2^ "i* A ^ = C013 a W32 .
In this case,
oJn A C032 = dtp a (— sin tp d&) = — sin tp dip a d&
which is the same as
cicoi2 = d(cosipd&) = d (cos ip) a dd 3= — sin <p d<p a rf#.
Sec. 8] THE STRUCTURAL EQUATIONS 95
Fi
p COS <p flW
FIG. 2.25
To derive the expressions given above for the dual 1 forms, first compute
dxi, dx2, dx 3 by differentiating the wellknown equations
Xi = p cos <p cos &
x 2 = p cos <p sin t?
x% = p sin <p.
Then substitute in the formula t = ^ a^ dx h where A = (a t y) is the
attitude matrix to be found in Example 6.2.
We shall later find a more efficient computational technique related to
the following reliable nonsense from elementary calculus: If at each point
the spherical coordinates p, #, <p are altered by dp, d&, d<p, then the sides of
the resulting infinitesimal box are dp, p cos <p d&, p d<p (Fig. 2.25). But these
are precisely the formulas for the dual forms 0i, 02, 3 .
As we mentioned earlier, our main use of the Cartan structural equations
is in studying the geometry of surfaces (Chapters VI and VII). A wider
variety of applications is given by Flanders [1].
EXERCISES
1. For a 1form <f> = ^2 fiQi, prove that
d<t> = E {dfj + L/wy) A 8j
i
(Compare Ex. 5 of II.7.)
2. For the toroidal frame field in Ex. 4 of II.6, show that
96 FRAME FIELDS [Chap. II
#i = dp 0012 = cos <p d&
2 — (R + p cos <p)d& a>i3 = d<p
03 = p d<p C023 = sin <p d&.
(Hint: Find 0* by the scheme described at the end of II.8. No computa
tions are necessary to find caul)
3. Check the first structural equations in the case of the toroidal frame
field.
4. For the cylindrical frame field E h E 2 , E 3 :
(a) Prove that 0i = dr, 2 = r d&, 3 = dz by evaluating on U u U 2 , U 3 .
(b) Deduce that EM = 1, E 2 [&] = 1/r, E 3 [z] = 1 and that the other
six possibilities Ei[&\, • • • are all zero.
(c) For a function /(r, t?, 0) expressed in terms of cylindrical coordi
nates, show that
m l Ewlg Bin 1
5. Frame fields on E 2 . For a frame field E u E 2 on the plane E 2 :
(a) Find the connection equations.
(b) If
Ei = cos <p Ui + sin <p U 2
E 2 = — sin (pUi \ cos <p U 2
where ^ is an arbitrary function, express the dual 1forms 0i, 2
and the connection form co i2 in terms of <p.
(c ) Prove the structural equations for this case.
9 Summary
We have accomplished the aims set at the beginning of this chapter. The
idea of a moving frame has been expressed rigorously as & frame field — either
on a curve in E , or on an open set of E itself. In the case of a curve, we
used only the Frenet frame field T, N, B of the curve. Expressing the
derivatives of these vector fields in terms of the vector fields themselves,
we discovered the curvature and torsion of the curve. It is already clear
that curvature and torsion tell a lot about the geometry of a curve; we
shall find in Chapter III that they tell everything. In the case of an open
set of E 3 , we dealt with an arbitrary frame field E u E 2 , E 3 . Cartan's gener
alization (Theorem 7.2) of the Frenet formulas followed the same pattern
of expressing the (covariant) derivatives of these vector fields in terms of
Sec. 9] SUMMARY 97
the vector fields themselves. Omitting the vector field V from the notation
in Theorem 7.2 we have
Cartan Frenet
V#i = CO12.EJ2 + COi3#3 T ' = KN
VEi = wiJ^i + W23E3 N' = kT + t£
Cartan's equations are not conspicuously more complicated than Fre
net 's because the notion of 1form is available for the coefficients &>,•/, the
connection forms.
CHAPTER III .
Euclidean Geometry
We recall some familiar features of plane geometry. First of all, two tri
angles are congruent if there is a rigid motion of the plane which carries
one triangle exactly onto the other. Corresponding angles of congruent
triangles are equal, corresponding sides have the same length, the areas
enclosed are equal, and so on. Indeed, any geometric property of a given
triangle is automatically shared by every congruent triangle. Conversely,
there are a number of simple ways in which one can decide if two given
triangles are congruent— for example, if for each the same three numbers
occur as lengths of sides.
In this chapter we shall investigate the rigid motions (isometries) of
Euclidean space, and see how these remarks about triangles can be extended
to other geometric objects.
1 Isometries of E 3
An isometry, or rigid motion, of Euclidean space is a special type of map
ping which preserves the Euclidean distance between points (Defini
tion 1.2, Chapter II).
1.1 Definition An isometry of E 3 is a mapping F: E 3 — ► E 3 such that
d(F(p),F(q)) =d(p,q)
for all points p, q in E 3 .
1.2 Example
(1) Translafions. Fix a point a in E 3 and let T be the mapping that adds
a to every point of E 3 . Thus T^p) = p + a for all points p. T is called
translation by a. It is easy to see that T is an isometry, since
98
Sec. 1]
ISOMETRIES OF E 3
99
FIG. 3.1
d(T(p),T(q)) = tf(p + a,q + a)
= (p + a)  (q + a)
= Up qll = d(p,q).
(2) Rotation around a coordinate axis. A rotation
of the xy plane through an angle # carries the
point (pi, p 2 ) to the point (51,92) with coordi
nates (Fig. 3.1)
Qi = Pi cos & — P2 sin #
#2 = Pi sin # + P2 cos #.
Thus a rotation C of E around the 2 axis (through an angle t?) has the
formula
C(p) = C (pi, p 2 , p 3 ) = (pi cos # — p 2 sin #, pi sin # + p 2 cos &, p 3 )
for all points p. Evidently, C is a linear transformation, hence, in particular,
a mapping. A straightforward computation shows that C preserves Euclid
ean distance, so C is an isometry.
Recall that if F and G are mappings of E 3 , the composite function GF
is a mapping of E 3 obtained by applying first F, then G.
1.3 Lemma If F and G are isometries of E 3 , then the composite mapping
GF is also an isometry of E 3 .
Proof. Since G is an isometry, the distance from G(F(p)) to (?(F(q))
is d(F(p), F(<i)). But since F is an isometry, this distance equals d(p, q).
Thus GF preserves distance, hence is an isometry. 
In short, a composition of isometries is again an isometry.
We also recall that if F: E 3 — » E 3 is both onetoone and onto, then F
has a unique inverse function F _1 : E 3 — > E 3 , which sends each point F(p)
back to p. The relationship between F and F~* is best described by the
formulas
FF~ l = I, F' X F = I
where / is the identity mapping of E 3 , that is, the mapping such that
^(p) = P f° r an P
The translations of E (as defined in Example 1.2) are the simplest
type of isometry.
1.4 Lemma
translation.
(1) If aS and T are translations, then ST = TS is also a
100 EUCLIDEAN GEOMETRY [Chap. Ill
(2) If T is translation by a, then T has an inverse T~\ which is trans
lation by — a.
(3) Given any two points p and q of E 3 , there exists a unique transla
tion T such that T(p) = q.
Proof. To prove (3), for example, note that translation by q  p cer
tainly carries p to q. This is the only possibility, since if T is translation
by a and T(p) = q, thenp + a = q; hence a = q  p. 
A useful special case of (3) is that if T is a translation such that for some
one point T(p) = p, then T = I.
The rotation in Example 1.2 is an example of an orthogonal transforma
tion of E , that is, a linear transformation C: E 3 — > E 3 which preserves
dot products in the sense that
C(p)»C(q) = p»q forallp,q.
1.5 Lemma If C: E 3 — > E 3 is an orthogonal transformation, then C is
an isometry of E 3 .
Proof. First we show that C preserves norms. By definition p  2 = p.p;
hence
IIC(p) 2 = C(p).c( P ) = P .p= p  2 .
Thus  C(p) = p  for all points p. Since C is linear, it follows easily
that C is an isometry:
d(C(p), C(q)) =  C(p)  C(q) =  C(p  q) = p  q.
= d(p, q) for all p, q. 
Our goal now is Theorem 1.7, which asserts that every isometry can
be expressed as an orthogonal transformation followed by a translation.
The main part of the proof is the following converse of Lemma 1.5.
1.6 Lemma If F is an isometry of E 3 such that ^(0) = 0, then F is
an orthogonal transformation.
Proof. First we show that F preserves dot products; then we show that
F is a linear transformation. Note that by definition of Euclidean distance,
the norm  p  of a point p is just the Euclidean distance d(0, p) from the
origin top. By hypothesis, F preserves Euclidean distance, and F(0) = 0;
hence
 F(p)\\ = d(0, F(p)) = d(F(0), F(p)) = d(0,p) = p .
Thus F preserves norms. Now by a standard trick ("polarization") we
shall deduce that it also preserves dot products. Since F is an isometry,
Sec. 1] ISOMETRIES OF E 3 101
d{F(p),F(q)) =rf(p,q)
for any pair of points. Hence
\\F(p)F( q )\\ = p q l,
which, by definition of norm, implies
(F(p)F(q)).(F(p) F(q)) = (pq)(pq).
Hence
 F(p)f  2F<p).F(q) +  F(q) 2 = p II 2  2p.q +  q  2 .
The norm terms here cancel, since F preserves norms, and we find
F(p)F(q) =p.q,
as required.
It remains to prove that F is linear. Let ui,u 2 ,u 3 be the unit points
(1, 0, 0), (0, 1, 0), (0, 0, 1), respectively. Then we have the identity
p = (pi,p2,pz) = 2 Pi^i
Also, the points ui,U2,U3 are orthonormal; that is, u,uy = bij.
We know that F preserves dot products, so F{u x ), F(u 2 ), F(u 3 ) must
also be orthonormal. Thus orthonormal expansion gives
F(p) = ZF^FMFiui).
But
F(p)F(ui) =p*iii = p h
so
F(p) = T,PiF( Ui ).
Using this identity, it is a simple matter to check the linearity condition
F(ap + 6q) = aF(p) + bF(q). 
We now give a concrete description of what an arbitrary isometry is like.
1.7 Theorem If F is an isometry of E 3 , then there exist a unique trans
lation T and a unique orthogonal transformation C such that
F = TC.
Proof. Let T be translation by F(0). We saw in Lemma 1.4 that T~ l
is translation by — F(0). But T~ l F is an isometry, by Lemma 1.3, and
furthermore,
(T 1 /0(0) = T~\F(0)) = F(0)  F(0) = 0.
102 EUCLIDEAN GEOMETRY [Chop. Ill
Thus by Lemma 1.6, T~*F is an orthogonal transformation, say T~*F = C.
Applying T on the left, we get F = TC.
To prove the required uniqueness, we suppose that F can also be ex
pressed as TC, where f is a translation and C an orthogonal transformation.
We must prove f = T and C = C. Now TC = TC; hence C = T~ l TC.
Since C and C are linear transformation, they of course send the origin to
itself. It follows thatjT _1 f)(0) = 0. But since T~ l f is a translation,
we conclude that T~'T = I; hence T = T. Then the equation TC = TC
becomes TC = TC. Applying T~\ we get C = C. 
Thus every isometry of E 3 can be uniquely described as an orthogonal trans
formation followed by a translation. When F = TC as in Theorem 1.7, we
call C the orthogonal part of F, and T the translation part of F. Note that
CT is generally not the same as TC (Exercise 1).
The decomposition theorem above is the decisive fact about isometries
of E 3 (and its proof holds for E" as well). For example, we shall now find
explicit formulas for an arbitrary isometry F = TC. If (c,y) is the ma
trix of the linear transformation C, we have the explicit formula
C(Pii P2, ps) = (S CijPj, H c*,Vh 2 c 3j pj)
for all points p = (pi,Pi,pz). We are using the columnvector conventions,
under which q = C(p) means
^A /Cll Ci 2 C13
?2 I = I Cn C 22 C 23
\W \c 3 i C S2 c M .
Since C is an orthogonal linear transformation, it is easy to show that its
matrix (dj) is orthogonal in the sense: inverse equals transpose.
Returning to the decomposition F = TC, suppose that T is translation
by a = (a u a 2 , a 3 ). Then
F(p) = TC(p) = a + C(p).
Using the above formula for C(p), we get
F(p) = F(p h p 2 ,p 3 ) = (oi + £ ciip h 02 + X) c 2 y?>i, a 3 + 2 <&#,■).
Alternatively, using the columnvector conventions, q = F(q) means
f Qi\ /oA / c " C 12 ci 3 \ /p^
g 2 J = I a 2 J + I c 2 i C22 c 2 3 I I P2
\qJ W/ Vsi c 32 C33/ \p 3 y
Sec. 1] ISOMETRIES OF E 3 103
EXERCISES
Throughout these exercises, A, B, and C denote orthogonal transforma
tions (or their matrices), and T a is translation by a.
1. Prove that CT a = T CW C
2. Given isometries F = T a A and G = T b B, find the translation and
orthogonal part of FG and GF.
3. Show that an isometry F = T a C has an inverse mapping F~ x , which is
also an isometry. Find the translation and orthogonal parts of F~ l .
4. If
C = [ i I f ] and ( P = , (3 ' *' ~ 6) '
1 2
3 3
q= (1,0,3)
show that C is orthogonal; then compute C(p) and C(q), and. check
thatC(p).C(q) =p.q.
5. Let F = T a C, where a = (1,3,1) and
/1/V2
C =
\1/V2
If p = (2, —2, 8), find the coordinates of the point q for which
(a)q = F(p). (b)q = ^(p). (c) q = (CT )(p).
6. In each case decide whether F is an isometry of E 3 . If so, find its trans
lation and orthogonal parts.
(a) F(j>) = p.
(b) ^(p) = p»a a, where  a  = 1.
(c) F(p) = (vz 1,P22, V1  3).
(d)F(p) = (vuP2,l).
A group G is a set furnished with an operation that assigns to each
pair g lf g 2 of elements of G an element <7i0 2 , subject to these rules: (1)
associative law: (gig^gs = ?ite), (2) there is a unique identity
element e such that eg = ge — g for all g in G, and (3) inverses: For
each g in G there is an element gr _1 in G such that gg~ l = g~ l g= e.
Groups occur naturally in many parts of geometry, and we shall men
tion a few in subsequent exercises. Basic properties of groups may be
found, for example, in Birkhoff and MacLane [2].
7. Prove that the set 8 of all isometries of E 3 forms a group — with compo
sition of functions as the operation. 8 is called the Euclidean group (of
order 3), or the group of Euclidean motions of E 3 .
104 EUCLIDEAN GEOMETRY [Chap. Ill
A subset H of a group G is a subgroup of G provided (1) if g x and g 2
are in H, then so is g^, (2) if g is in H, so is gr\ and (3) the identity
element e of G is in #. A subgroup H of G is automatically a group.
8. Prove that the set 3 of all translations of E 3 and the set 0(3) of all
orthogonal transformations of E 3 are each subgroups of the Euclidean
group S. (3) is called the orthogonal group of order 3. Which isometries
of E 3 are in both these subgroups?
2 The Derivative Map of an Isometry
In Chapter I we showed that an arbitrary mapping F: E 3 *• E 3 has a
derivative map F* which carries each tangent vector v at p to a tangent
vector F*(\) at F(p). If F is an isometry, its derivative map is remarkably
simple. (Since the distinction between tangent vector and point is crucial
here, we temporarily restore the point of application to the notation.)
2.1 Theorem Let F be an isometry of E 3 with orthogonal part C. Then
n(v p ) = (CVW,
for all tangent vectors \ p to E 3 .
Verbally: To get F+(v p ), first shift the tangent vector v p to the canoni
cally corresponding point v of E 3 , then apply the orthogonal part C of F,
and finally shift this point C(v) to the canonically corresponding tangent
vector at F(p) (Fig. 3.2). Thus all tangent vectors at all points p of E 3 are
"rotated" in exactly the same way by F* — only the new point of application
F(p) depends on p.
Proof. Write F = TC as in Theorem 1.7. Let T be translation by a,
so F(p) = a + C(p). If \ p is a tangent vector to E 3 , then by Definition
7.4 of Chapter I, F*(v p ) is the initial velocity of the curve t — > F(p + tv).
But using the linearity of C, we obtain
F(p + tv) = TC(p + tv) = T(C(p) + tC(v)) = a + C(p) + tC(v)
= F(p) + tC(v).
FIG. 3.2
Sec. 2] THE DERIVATIVE MAP OF AN ISOMETRY 105
Thus jP*(v p ) is the initial velocity of the curve t — > F(p) + tC(v), which
is precisely the tangent vector (Cv) F(p ). 
Expressed in terms of Euclidean coordinates, this result becomes
F*(E, VjUj) = X) CijVjUi
i i,i
where C = (dj) is the orthogonal part of the isometry F, and if Ui is
evaluated atp, then Ui is evaluated at F(p).
2.2 Corollary Isometries preserve dot products of tangent vectors.
That is, if \ p and w p are tangent vectors to E at the same point, and F
is an isometry, then
F*(v p )»F*(w p ) = \ p 'w p .
Proof. Let C be the orthogonal part of F, and recall that C, being an
orthogonal transformation, preserves dot products in E 3 . By Theorem 2.1,
F*{y,)*F*(w,) = (Cv) np) (Cw) F(p) = CyCw
= VW = Vp'Wp
where we have twice used Definition 1.3 of Chapter II (dot products of
tangent vectors). 
By proving this fundamental corollary and the following theorem, the
initial result (Theorem 2.1) has largely accomplished its mission. Thus
it should be safe to drop the point of application from the notation once
again, and write simply F*(\)»F*(w) = vw. In fancier language, the
corollary asserts that for each point p, the derivative map F* p at p is an
orthogonal transformation of tangent spaces (differing from C only by
the canonical isomorphisms of E 3 ).
Since dot products are preserved, it follows automatically that derived
concepts such as norm and orthogonality are preserved. Explicitly, if F
is an isometry, then  /^*Cv) =  v [, and if v and w are orthogonal, so
are jP*(v) and F*(w). Thus frames are also preserved: if ci, e 2 , e 3 is a frame
at some point p of E 3 and F is an isometry, then / r *(e 1 ), F*(e 2 ), F*(e 3 )
is a frame at F(p). (A direct proof is easy: e^ey = 8^, so by Corollary 2.2,
F*(e<).F*(c y ) = ere, = 5< y .)
Assertion (3) of Lemma 1.4 shows how two points uniquely determine a
translation. We now show that two frames uniquely determine an isometry.
2.3 Theorem Given any two frames on E 3 , say d, e^, e 3 at the point p
and fi, f 2 , f 3 at the point q, there exists a unique isometry F of E 3 such that
F»fe<) = f,forl ^ i£ 3.
Proof. First we show that there is such an isometry. Let ei, e 2 , e 3 , and
/i, h, h be the points of E canonically corresponding to the vectors in
106 EUCLIDEAN GEOMETRY [Chap. Ill
the two frames. Let C be the unique linear transformation of E 3 such that
C(e») = fi for 1 ^ i ^ 3. It is easy to check that C is orthogonal. Then
let T be a translation by the point q  C (p). Now we assert that the isome
try F = TC carries the e frame to the f frame. First note that
F(p) = T(Cp) = q  C(p) + C(p) = q.
Then using Theorem 2.1 we get
F*(e t ) = (Ce t ), (p) = (/.),(,) = (/„■), = f {
for 1 ^ i ^ 3.
To prove uniqueness, we observe that by Theorem 2.1 this choice of C
is the only possibility for the orthogonal part of the required isometry.
The translation part is then completely determined also, since it must carry
C(p) to q. Thus the isometry F = TC is uniquely determined. 
Explicit computation of the isometry in the theorem is not difficult.
Let e, = (a a , a i2 , o«) and /< = (6 a , b a , b a ) for 1 ^ i ^ 3. Thus the (or
thogonal) matrices A = (a ti ) and £ = (6 l7 ) are the attitude matrices of
the frames e x , es, e 3 and fi, f 2 , f 3 , respectively. We claim that C in the theo
rem (or, strictly speaking, its matrix) is l BA. It suffices to show that
BA (ei ) = /,, since this uniquely characterizes C. But using the column
vector conventions, we have
'BA a 12 = '£ U
\aj W \6 13/
that is, 'BA(ei) = /i (the cases i = 2, 3 are similar). Thus C = l BA
As noted above, 7 1 is then necessarily translation by q — C(p).
EXERCISES
1. If T is a translation, then for every tangent vector v show that 7 1 *(v)
is parallel to v (same Euclidean coordinates).
2. Prove the general formulas (GF)* = G*F* and (F 1 )* = (f 7 *) 1
in the special case where F and G are isometries of E 3 .
3. (a) Let e x , e2, e 3 be a frame at p with attitude matrix A. If F is the
isometry that carries the natural frame at to this frame, show
thatF = T p A~ l (A 1 = l A).
(b) Now let fi, f 2 , f 3 be a frame at q with attitude matrix B. Use Ex
ercise 2 to prove the result in the text that the isometry which
carries the e frame to the f frame has orthogonal part B _1 A.
Sec. 3] ORIENTATION 107
4. (a) Prove that an isometry F = TC carries the plane through p or
thogonal to q to the plane through F(p) orthogonal to C(q).
(b) If P is the plane through (£, — 1,0) orthogonal to (0, 1,0) find an
isometry F = TC such that F(P) is the plane through (1, 2, 1)
orthogonal to (1 , 0, — 1 ) .
5. Given the frame ei = (2,2,l)/3, e 2 = (2,l,2)/3, e 3 = (1, 2,2)/3
at p = (0, 1,0) and the frame
f 2 = (l,0,l)/\/2, f 2 = (0,1,0), f 3 = (1,0, 1)/V2
at q = (3, —1, 1), find the isometry F = TC which carries the
e frame to the f frame.
3 Orientation
We now come to one of the most interesting and elusive ideas in geometry.
Intuitively, it is orientation that distinguishes between a righthanded glove
and a lefthanded glove in ordinary space. To handle this concept mathe
matically, we replace gloves by frames, and separate all the frames on E 3
into two classes as follows. Recall that associated with each frame ei, e^, e 3
at a point of E is its attitude matrix A . According to the exercises for Sec
tion 1 of Chapter II,
ei»e 2 X e 3 = det A = ±1.
When this number is +1, we shall say that the frame e x , ej, e 3 is positively
oriented (or, righthanded); when it is —1, the frame is negatively oriented
(or, lefthanded).
We omit the easy proof of the following facts.
3.1 Remark (1) At each point of E 3 the frame assigned by the natural
frame field U u £/ 2 , U 3 is positively oriented.
(2) A frame ei, e%, e 3 is positively oriented if and only if e x X ^ = e 3 .
Thus the orientation of a frame can be determined, for practical purposes,
by the "righthand rule" given at the end of Section 1 of Chapter Il.Pic
torially, the frame (P) in Fig. 3.3 is positively oriented, whereas the frame
(N) is negatively oriented. In particular, Frenet frames are always positively
oriented, since by definition B = T X N.
(3) For a positively oriented frame e x , e 2 , e 3 , the cross products are
ei = e 2 X e 3 = — e 3 X ^
e 2 = e 3 X ei = — e x X e 3
e 3 = ei X e 2 = — e 2 X ei.
For a negatively oriented frame, reverse the vectors in each cross product.
108
EUCLIDEAN GEOMETRY
[Chap. Ill
(P)
e 2
FIG. 3.3
(N)
ei
(One need not memorize these formulas — the righthand rule will give
them all correctly.)
Having attached a sign to each frame on E 3 , we next attach a sign to
each isometry F of E 3 . In Chapter II we proved the wellknown fact that
the determinant of an orthogonal matrix is either +1 or —1. Thus if C is
the orthogonal part of the isometry F, we define the sign of F to be the
determinant of C, with notation
sgn F = det C.
We know that the derivative map of an isometry carries frames to
frames. The following result tells what happens to their orientations.
3.2 Lemma If e x , e2, e 3 is a frame at some point of E 3 and F is an isome
try, then
F*(c0.F*(©2) X /^*(e 3 ) = sgnF ei .e 2 X e 3 .
Proof. If ej = 23 a JkU k , then by the coordinate form of Theorem 2.1
we have
F*(ej) = z2 CikdjkUi
i,k
where C = (cij) is the orthogonal part of F. Thus the attitude matrix of
the frame / p *(e 1 ), F^M, ^*(e 3 ) is the matrix
(2 cafljk) = (23 c ik Wy) = C l A.
k k
But the triple scalar product of a frame is the determinant of its attitude
matrix, and by definition sgn F = det C. Consequently,
F*( ei ).F*(e 2 ) X i^M = det (C l A)
= det (7det *A = det Cdet A
= sgnjPei«e2 X e 3 .
I
This lemma shows that if sgn F = +1, then F* carries positively oriented
frames to positively oriented frames, and carries negatively oriented to
Sec. 3]
ORIENTATION
109
negatively oriented frames. On the other hand, if sgn F — — 1, positive
goes to negative, and negative to positive.
3.3 Definition An isometry F of E 3 is said to be
orientationpreserving if sgn F = det C = + 1
orientationreversing if sgn F = det C = — 1
where C is the orthogonal part of F.
3.4 Example
(1) Translations. All translations are orientationpreserving. Geometrically
this is clear, and in fact the orthogonal part of a translation T is just the
identity mapping i", so sgn T = det / = +1.
(2) Rotations. Consider the orthogonal transformation C given in Example
1.2, which rotates E 3 through angle around the z axis. Its matrix is
COS0
— sin
sin 6
cos 6
1
Hence sgn C — det C = +1, so C is orientationpreserving (see Exercise 4).
(3) Reflections. One can (literally) see reversal of orientation by using a
mirror. Suppose the yz plane of E 3 is the mirror. If one looks toward that
plane, the point p = (pi,P2,P3) appears to be located at the point
R(P) = (~PhP2,Ps)
(Fig. 3.4). The mapping R so denned is called reflection in the yz plane.
Evidently it is an orthogonal transformation with matrix
1
1
1
Thus R is an orientationreversing isom
etry, as confirmed by the experimental
fact that the mirror image of a right hand
is a left hand.
Both dot and cross product were orig
inally defined in terms of Euclidean co
ordinates. We have seen that the dot
product is given by the same formula,
VW = (52 Vi«i)" (22 «\«t) = 52 v Wi,
,<°R(p)
edge view of yz plane
FIG. 3.4
ei
e 2
e 3
Vi
V2
Vi
W\
102
w 3
110 EUCLIDEAN GEOMETRY [Chap. Ill
no matter what frame ei, e^ e 3 is used to get coordinates for v and w.
We have almost the same result for cross products, but orientation is now
involved.
3.5 Lemma Let d, 62, e 3 be a frame at a point of E 3 . If v = ^ v&i
and w = ^2 Wiei, then
v X w = e
where e = e^^ X e 3 = ±1.
Proof. It suffices merely to expand the cross product
v X w = (wid + w 2 e 2 + v 3 e 3 ) X (wiei + W2e<s + w 3 e 3 )
using the formulas (3) of Remark 3.1. For example, if the frame is posi
tively oriented, for the ei component of v X w, we get
the* X w 3 e 3 + v 3 e 3 X W2e2 = (v 2 w 3 —v 3 w 2 )ei.
Since e = 1 in this case, we get the same result from the right side of the
equation to be proved. 
It follows immediately that the effect of an isometry on cross products
also involves orientation.
3.6 Theorem Let v and w be tangent vectors to E at p. If F is an isOme
try of E 3 , then
F*(v X w) = sgn F F*(v) X F*(w).
Proof. Write v = ^ i>,t/,(p) and w = YL WiUifo). Now let
e, = F+(Ui(p)).
Since F* is linear,
F*(y) = 2 v i e i an d F*(w) = 2 «*,«,•.
A straightforward computation using Lemma 3.5 shows that
F*(v) X F*(w) = e F*(v X w),
where
e = ei .e 2 X e 3 = F,(C7i(p)).F* (C7 2 (p)) X F*(t/ 3 (p)).
But t/i, C7 2 , U 3 is positively oriented, so by Lemma 3.2, e = sgn F. 
Sec. 3]
ORIENTATION
111
EXERCISES
1. Prove
sgn (FG) = sgn Fsgn G = sgn (GF).
Deduce that sgn F = sgn (F _1 ).
2. If Ho is an orientationreversing isometry of E 3 , show that every orien
tationreversing isometry has a unique expression HqF, where F is
orientationpreserving.
3. Let v = (3, 1, —1) and w = (3, 3, 1) be tangent vectors at some
point. If C is the orthogonal transformation given in Exercise 4 of
Section 1, check the formula
C*(v X w) = sgn C C*(v) X C*(w).
4. A rotation is an orthogonal transformation C such that det C = +1.
Prove that C does, in fact, rotate E 3 around an axis. Explicitly, given a
rotation C, show that there exists a number # and points d, d, e 3 with
e t »ey = Sij such that (Fig. 3.5)
C (ei ) = cos # d + sin # e 2
C(e 2 ) = —sin i? d + cos # d
C(e.) = e 3 .
(Hint: The fact that the dimension of E
is odd means that C has a characteristic
root + 1, so there is a point p ^ such that
C(p) =p.)
5. Let a be a point of E 3 such that  a  = 1. Prove that the formula
C(p) = a X p +p*a a
defines an orthogonal transformation. Describe its general effect on E .
6. Prove
(a) The set + (3) of all rotations of E 3 is a subgroup of the orthogonal
group 0(3) (see Ex. 8 of III.l).
(b) The set 8 + of all orientationpreserving isometries of E is a subgroup
of the Euclidean group S.
7. Find a single formula for all isometries of the real line E 1 . Do the same
for the plane E 2 (use e = ±1). Which of these isometries are orienta
tionpreserving?
FIG. 3.5
1,2 EUCLIDEAN GEOMETRY [Chap. Ill
4 Euclidean Geometry
In the discussion at the beginning of this chapter, we recalled a fundamental
feature of plane geometry: If there is an isometry carrying one triangle
onto another, then the two (congruent) triangles have exactly the same
geometric properties. A close examination of this statement will show
that it does not admit a proof — it is, in fact, just the definition of "geo
metric property of a triangle." More generally, Euclidean geometry can be
defined as the totality of concepts that are preserved by isometries of
Euclidean space. For example, Corollary 2.2 shows that the notion of dot
product on tangent vectors belongs to Euclidean geometry. Similarly
Theorem 3.6 shows that the cross product is preserved by isometries (ex
cept possibly for sign).
This famous definition of Euclidean geometry is somewhat generous,
however. In practice, the label "Euclidean geometry" is usually attached
only to those concepts that are preserved by isometries, but not by arbi
trary mappings, or even the more restrictive class of mappings (diffeo
morphisms) that possess inverse mappings. An example should make this
distinction clearer. If a = (ai,a 2 ,a 3 ) is a curve in E 3 , then the various
derivatives
a = ( ^ gl ^? d<*z \ „ _ (d 2 ai d 2 a 2 d 2 <x 3 \
look pretty much alike. Now, we interpreted Theorem 7.8 of Chapter I
as saying that velocity is preserved by arbitrary mappings T^: E 3 — >■ E 3 . That
is, if = F(a), then /? = F*(a ). But it is easy to see that acceleration
is not preserved by arbitrary mappings. For example, if a(t) = (2,0,0) and
F = (x\y,z), then a" = 0; hence F+{a") = 0. But p = F(a) has the
formula 0(0 = (* 2 ,0,0), so 0" = 2C/i. Thus in this case, = F(a), but
$" 5^ F*(a"). We shall see in a moment, however, that acceleration is
preserved by isometries.
For this reason, the notion of velocity belongs to the calculus of Euclid
ean space, while the notion of acceleration belongs to Euclidean geometry.
In this section we examine some of the concepts introduced in Chapter II
and prove that they are, in fact, preserved by isometries. (We leave largely
to the reader the easier task of showing that they are not preserved by
diffeomorphisms. )
Recall the notion of vector field on a curve (Definition 2.2 of Chapter
II). Now if Y is a vector field on a: I > E 3 , and F: E 3 — ► E 3 is any mapping,
then Y = F*(Y) is a vector field on the image curve a = F(a). In fact,
for each /in/, Y(t) is a tangent vector to E 3 at the point a (t). But then
Y(t) = F*(Y(t)) is a tangent vector to E 3 at the point F(a(t)) = a(t).
Sec. 4]
EUCLIDEAN GEOMETRY
113
FIG. 3.6
(These relationships are illustrated in Fig. 3.6.) Isometries preserve the de
rivatives of such vector fields.
4.1 Corollary Let Y be a vector field on a curve a in E 3 , and let F be
an isometry of E 3 . Then Y = F*(F) is a vector field on a = F(a), and
f = F+iY*).
Proof. We compute F*(Y') and Y starting from the expression
Y = Z ViUt
for Y in terms of its Euclidean coordinate functions. To differentiate such
a vector field Y, one simply differentiates its Euclidean coordinate functions,
so
F' = Zf V,
Thus by the coordinate version of Theorem 2.1, we get
On the other hand,
F*(Y') =T,^U i .
Y = F*(Y) = T.CiiViUi.
But each c t _, is constant, being by definition an entry in the matrix of the
orthogonal part of the isometry F. Hence
Thus the vector fields F*(Y') and Y' are the same.
I
We claimed earlier that isometries preserve acceleration: If a = F(a),
where F is an isometry, then a" = F*(a"). This is an immediate conse
"4 EUCLIDEAN GEOMETRY [Chap. Ill
quence of the preceding result, for if we set Y = a, then by Theorem 7.8
of Chapter I, Y = a; hence
5 " =Y' = F*(Y') = F*(aT).
Now we show that the Frenet apparatus of a curve is preserved by
isometries. This is certainly to be expected on intuitive grounds, since a
rigid motion ought to carry one curve into another that turns and twists
in exactly the same way. And this is what happens when the isometry is
orientationpreserving .
4.2 Theorem Let be a unitspeed curve in E 3 with positive curvature,
and let = F(0) be the image curve of under an isometry F of E 3 . Then
k = k f = F*(T)
f = sgn Ft N = F*(N)
B = sgn FF*(B)
where sgn F = ±1 is the sign of the isometry F.
Proof. Note that is also a unitspeed curve, since
ii in = ii^goii = \\ft\\ = i.
Thus the definitions in Section 3 of Chapter II apply to both and /3, so
f = ff = F*(/3') = F m (T).
Since F* preserves both acceleration and norms, it follows from the
definition of curvature that
* = H01 = imonn = ii0" ii = *■
To get the full Frenet frame, we now use the hypothesis k > (which
implies Z > 0, since R = k). By definition, N = fi"/n; hence using preceding
facts, we find
K K \ K /
It remains only to prove the interesting cases B and t. Since the defini
tion B = T X N involves a cross product, we use Theorem 3.6 to get
B = f X N = F*(T) X F*(N) = sgn FF*(T X N) = 8gnFF*(B).
The definition of torsion is essentially r = —B'»N = BN'. Thus using
the results above for B and N, we get
f = B.N' = sgn F F*(B)*F*(N') = sgn F BN' = sgn Ft. 
The presence of sgn F in the formula for the torsion of F{fi) shows that
Sec. 4]
EUCLIDEAN GEOMETRY
115
the torsion of a curve gives a more subtle description
of the curve than has been apparent so far. The sign
of t measures the orientation of the twisting of the
curve. If F is orientationreversing, the formula f =
— t proves that the twisting of the image of curve
F(fi) is exactly opposite to that of itself.
A simple example will illustrate this reversal.
4.3 Example Let /3 be the unitspeed helix
8(s) = (cos , sin  , ),
\ c c c/
FIG. 3.7
gotten from Example 3.3 of Chapter II by setting
a = b = 1 ; hence c = \/2 We know from the general formulas for helices
that k = t = \. Now let R be reflection in the xy plane, so R is the isom
etry R{x,y,z) = (x,y,—z). Thus the image curve (8 = R(p) is the mirror
image
/8(«) =
(s . s s\
cos  , sin  , —  J
c c c/
of the original curve. One can see in Fig. 3.7 that the mirror has its usual
effect: and fi twist in opposite ways — if /3 is "righthanded," then /3 is
"lefthanded." (The fact that is going up and j§ down is, in itself, irrele
vant.) Formally: The reflection R is orientationreversing; hence the
theorem predicts k = k = \ and f = — t = —\. Since /8 is just the helix
gotten in Example 3.3 of Chapter II by taking a = 1 and b = — 1, this
may be checked by the general formulas there.
EXERCISES
1. Let F = TC be an isometry of E 3 , /3 a unit speed curve in E 3 . Prove
(a) If /3 is a cylindrical helix, then F(f3) is a cylindrical helix.
(b) If /3 has spherical image ft, then F(fi) has spherical image C(j8).
2. Let Y = (t, 1 — t 2 , 1 + t 2 ) be a vector field on the helix
a(t) = (cos t, sin t, 2t),
and let C be the orthogonal transformation
1
C =  1A/2 1/V2
1/V2 1/V2/
"* EUCLIDEAN GEOMETRY [Chap. Ill
Compute a = C(a) and Y = C*(F), and check that
C*(F / ) = ?', C*(a") = a", F'a" = ?'.«".
3. Sketch the triangles in E 2 that have vertices
A i: (3, 1), (7, 1), (7, 4) A 2 : (2, 0), (2, 5), (f, ^)
Show that these triangles are congruent by exhibiting an isometry F
that carries Ai to A 2 . (Hint: the orthogonal part of F is not altered if
the triangles are translated.)
4. If F: E —> E 3 is a mapping such that F* preserves dot products, show
that F is an isometry. (Hint: Use Ex. 11 of II.2.)
5. Let F be an isometry of E 3 . For each vector field V let V be the vector
field such that F*( F(p) = 7(F(p) ) for all p. P rove that isometries pre
serve covariant derivatives; that is, show y v W = VrW.
5 Congruence of Curves
In the case of curves in E 3 , the general notion of congruence takes the
following form.
5.1 Definition Two curves a ,0: I — > E are congruent provided there
exists an isometry F of E 3 such that ft = F(a); that is, /?(£) = F(a(t))
for all tin I.
Intuitively speaking, congruent curves are the same except for position
in space. They represent trips at the same speed along routes of the same
shape. For example, the helix a(t) = (cos t, sin t, t) spirals around the
z axis in exactly the same way the helix (i(t) = (t, cos t, sin t) spirals
around the x axis. Evidently these two curves are congruent, since if F
is the isometry such that
F(Vu Pi, P») = (PfcPi, P*),
then F (a) = p.
To decide whether given curves a and /8 are congruent, it is hardly
practical to try all the isometries of E 3 to see if there is one that carries
a to 8. What we want is a description of the shape of a unitspeed curve so
accurate that if a and /? have the same description, then they must be
congruent. The proper description, as the reader will doubtless suspect,
is given by curvature and torsion. To prove this we need one preliminary
result.
Curves whose congruence is established by a translation are said to be
parallel. Thus curves a, /3: / — * E are parallel if and only if there is a point
Sec. 5]
CONGRUENCE OF CURVES
117
p in E 3 such that /3(s) = a(s) + p for all s in /, or, in functional nota
tion, /3 = a + p.
5.2 Lemma Two curves a,/3: / — »• E 3 are parallel if their velocity vectors
a (s) and (s) are parallel for each s in /. In this case, if a(so) = P(s )
for some one s in /, then a = 0.
Proof. By definition, if a (s) and /8 (s) are parallel, they have the same
Euclidean coordinates. Thus
— f ") = — M
for 1 < i < 3
where a» and /3i are the Euclidean coordinate functions of a and /S. But
by elementary calculus, the equation doti/ds = dPi/ds implies that there is a
constant pi such that ft = a t + Pi Hence /3 = a + p. Furthermore, if
a(so) = (8(so), we deduce thatp = 0; hence a = 0. 
5.3 Theorem If a, /3: / — » J57 are unitspeed curves such that *c« = k#
and r a = ±70, then a and /3 are congruent.
Proof. There are two main steps:
(1) Replace a by a suitably chosen congruent curve F(a).
(2) Show that F(a) = 9 (Fig. 3.8).
Our guide for the choice in (1 ) is Theorem 4.2. Fix a number, say 0,
in the interval /. If r a = t/j, then let F be the (orientationpreserving)
isometry that carries the Frenet frame T a (0), N a (0), B a (0) of a at a (0)
to the Frenet frame T(0), N(0), B(0), of ■ at /3(0). (The existence of this
isometry is guaranteed by Theorem 2.3.) Denote the Frenet apparatus of
a = F(a) by k, f, f, N, B; then it follows immediately from Theorem 4.2
and the information above that
«(0) = 0(0)
K = Kp
f = 70
«m
T(0)
= T(0)
N(0)
= N(0)
8(0)
= B(0).
a = F{aV)
4
(t)
5(0) = /3(0>
FIG. 3.8
118 EUCLIDEAN GEOMETRY [Chap. Ill
On the other hand, if r a = —t$, we choose F to be the (orientation
reversing) isometry that carries T a (0), N a (0), B a (0) at a(0) to the frame
T(0), N(0), —B(0) at /3(0). (Frenet frames are positively oriented;
hence this last frame is negatively oriented : This is why F is orientation
reversing.) Then it follows from Theorem 4.2 that the equations (f) hold
also for a = F(a) and £. For example,
B(0) = F*(B a (0)) = 5(0).
For step (2) of the proof, we shall show T = T; that is, the unit tangents
of a = F(a) and are parallel at each point. Since a(0) = /3 (0 ) , it will
follow from Lemma 5.2 that F(a) = /?. On the interval /, consider the
real valued function/ = T»T + N»N + B'B. Since these are unit vector
fields, the Schwarz inequality (Section 1, Chapter II) shows that
TT g: 1;
furthermore f »T = 1 if and only if f = T. Similar remarks hold for the
other two terms in /. Thus it suffices to show that f has constant value 3.
By (t),/(0) = 3. Now consider
/ = T'T + TT' + N'N + NN' + B'B + BB'.
A simple computation completes the proof. Substitute the Frenet formu
las in this expression and use the equations H = k, f = t from (J). The
resulting eight terms cancel in pairs, so / =0, and / has, indeed, constant
value 3. 
Thus a unitspeed curve is determined but for position in E by its curvature
and torsion.
Actually the proof of Theorem 5.3 does more than establish that a and /3
are congruent; it shows how to compute explicitly an isometry carrying
a to )8. We illustrate this in a special case.
5.4 Example Consider the unitspeed curves a, /3: R — » E such that
( \ ( s . s s\
a\s) = I cos  , sin ,  ]
\ c c c/
/3(s) = I cos  , sin  , )
\ c c c/
where c = \/2. Obviously these curves are congruent by means of a
reflection — they are the helices considered in Example 4.3 — but we shall
ignore this in order to describe a general method for computing the re
quired isometry. According to Example 3.3 of Chapter II, a and /8 have the
same curvature, n a = \ = Kp; but torsions of opposite sign, r a = \ = — 7>
Thus the theorem predicts congruence by means of an orientationrevers
Sec. 5] CONGRUENCE OF CURVES 119
ing isometry F. From its proof we see that F must carry the Frenet frame
T a (0) = (0,a, a)
N a (0) = (1,0,0)
B a (0) = (0, a, a)
to the frame
T,(0) = (0, a, a)
N (O) = (1,0,0)
B (O) = (0, a, a)
where a = l/\/2. (These explicit formulas also come from Example 3.3 of
Chapter II.) By the remark following Theorem 2.3, the isometry F has
orthogonal part C = l BA, where A and B are the attitude matrices of the
two frames above. Thus
a a\ /l 0\
1 0=01
a a/ \0 l y
since a = l/\/2 These two frames have the same point of application
a(0) = 8(0) = (1, 0, 0). But C does not move this point, so the transla
tion part of F is just the identity map. Thus we have (correctly) found
that the reflection F = C carries a to 0.
From the viewpoint of Euclidean geometry, two curves in E 3 are "the
same" if they differ only by an isometry of E 3 . What, for example, is a helix?
Not just a curve that spirals around the z axis as in Example 3.3 of Chapter
II, but any curve congruent to one of these special helixes. One can give
general formulas, but the best characterization follows.
5.5 Corollary Let a be a unit speed curve in E 3 . Then a is a helix if and
only if both its curvature and torsion are nonzero constants.
Proof. For any numbers a > and 6^0, let &,& be the special helix
given in Example 3.3 of Chapter II. If a is congruent to S„, 6 , then (changing
the sign of b if necessary) we can assume the isometry is orientation
preserving. Thus, a has curvature and torsion
a 2 + b 2 a 2 + 6 2 "
Conversely, suppose a has constant nonzero k and t. Solving the pre
ceding equations, we get
120 EUCLIDEAN GEOMETRY [Chap. Ill
* 1,
a = 7— — „ b =
K 2 + T 2 K 2 + T 2
Thus a and &,& have the same curvature and torsion, hence they are
congruent. 
Our results so far demand unit speed, but it is easy to weaken this re
striction.
5.6 Corollary Let a, /3: / — > E be arbitraryspeed curves. If
Va = Vp > 0, K a = K0 > 0, and T a — ±T0,
then the curves a and /3 are congruent.
Proof. Let 5 and $ be unitspeed reparametrizations of a and (3, both
based at, say, t = 0. Since a and jS have the same speed function, it follows
immediately that they also have the same arc length function s = s(t)
and hence the same inverse function t = t(s). But since
Ka = K(S and T a = ± T0,
we deduce from the general definitions of curvature and torsion in that Sec
tion 4 of Chapter II
Ka(s) = K a (t(s)) = Kfi(t(s)) = K$(s)
Ta(s) = T a (t(s)) = ±Tfi(t(s)) = ±T^(s).
Hence the congruence theorem (5.3) shows that a and are congruent
— say, F(a) = j8. But then the same isometry carries a to /?, since
F(a(t)) = F(a(s(t))) = F(p(s(t))) = F(p(t)). 
The theory of curves we have presented applies only to regular curves
with positive curvature, because only for such curves is it generally pos
sible to define the Frenet frame field. However, a completely arbitrary
curve a in E can be studied by means of an arbitrary frame field on a,
that is, any three unitvector fields Ei, E 2 , E 3 on a that are orthogonal at
each point (Fig. 3.9).
FIG. 3.9
Sec. 5] CONGRUENCE OF CURVES 121
For example, the congruence theorem 5.3 can easily be extended to arbi
trary curves.
5.7 Theorem Let a, /?: / — > E 3 be arbitrary curves, and let E h E 2 , E 3 be
a frame field on a; F u F 2 , F 3 a frame field on 0. If
(1) a''Ei = p'Fi (l£i ^ 3)
(2) EiEi = F/.Fj (1 ^ i,j S 3),
then a and are congruent.
Proof. We need only generalize the argument in Theorem 5.3. Fix a
number, say 0, in /. Then let F be the isometry that carries
E 1 (0), E 2 (0), E 3 (0) to F t (0), F 2 (0), F 3 (0).
Since F* preserves dot products, it follows that Ei = F*(Ei) (1 ^ i ^ 3)
is a frame field on a = F(a). Since F* preserves velocities and derivatives
of vector fields as well, we deduce
a(0) = 0(0) a' Ei = fl'Fi
Ei(0) = F,(0) E/.Ej = Ft'Fj for 1 ^ i,j ^ 3.
This last equation means we can write £/ = 23 QijEj and F/ = 23 G^i
with the same coefficient functions a,,. Note that a ti f a>i = 0. (Differ
entiate Ei*Ej = 8ij.) If / = 23 Ei'Fi, we then prove/ = 3 just as before,
since
/' = 23 (^/*\ + Ei'Fi') = 23 («<y + aii)E r Fi = 0.
Thus Ei = Fi (parallelism!) and it follows from ({) that
«' = E a'.E, & and /?' = 23 p'.Fi F {
are parallel at each point. But a(0) = /3(0); hence by Lemma 5.2,
F(a) = a = 8. 
We shall need this degree of generalization in Section 8 of Chapter VI.
A more elegant (but slightly less general) version of this theorem is given
in Exercise 3.
EXERCISES
1. Given a curve a = (a 1} a 2 , a 3 ): I — > E 3 , prove that 0: / — » E 3 is con
gruent to a if and only if /3 can be written
0(0 = P + ati(t)ei + a 2 (0e 2 + a 3 (t)e 3 ,
where e»»Cj = 5i_,.
122 EUCLIDEAN GEOMETRY [Chap. Ill
2. Let a be the curve in Example 4.4 of Chapter II. Find a (congruent)
curve of the form 7(0 = (at, bt , ct ) and an isometry F such that
F(a) = y.
3. Let Ei, E 2 , E3 be a frame field on E with dual forms 0; and connection
forms a>ij. Prove that two curves a, /3: / — » E are congruent if we have
Bi(a) = Oi(fi') and «,,■(«') = coaifi') for 1 ^ ij ^ 3.
4. Show that the curve
p(t) = (t + V3 sin t, 2 cos t, y/Z t  sin t)
is a helix by computing its curvature and torsion. Find a helix a of
the form (a cos t, a sin £, 60 and an isometry F such that F(a) = /3.
5. Let a,(i: I — *■ E be congruent curves with k > 0. Show that there is
only one isometry F such that F(a) = (8 — unless r = 0, in which case
there are exactly £wo.
6. (Continuation). Find the two isometries carrying the parabola a(t) =
(v%, £ 2 , 0) to the parabola 0(t) = (t,t,t 2 ).
7. If j8 is a unitspeed curve in E , then every unitspeed reparametriza
tion j8 of j8 has the form j8(s) = j8(±s + s ). If (3 and 8 are congruent,
this represents a symmetry in the common route of /3 and j§. Prove that
helical routes are completely symmetric. Explicitly, show that the
helix 8 in Example 3.3 of Chapter II is congruent to every unitspeed
reparametrization /3 by explicitly finding the isometry F = TC such
thatF(0) = /3.
8. Two curves a : I — * E and £ : / — » E have congruent routes provided
there is an isometry F such that F(a) is a reparametrization of 0.
(a) Show that unitspeed curves a and j8 have congruent routes if
and only if there is a number s such that K a (s) = Kp(es + so) and
t„(s) = dzTp(es + So), where e is either +1 or —1.
(b) If a is the curve in Ex. 2 of II.4 show a and = (V, e~'/2,
have congruent routes. Exhibit the isometry F = TC and the re
parametrization needed to fulfill the definition.
The following three exercises deal with curves in E 2 .
9. Given any differentiate f function k on an interval /, prove that there
is a unitspeed curve a in E 2 such that k is the curvature function of a.
(Hint: Find an integral formula for a by reversing the order of results
in Ex. 8 of II.3.)
10. Find plane curves — in any convenient parametrization — for which
t Even if k is merely continuous, we obtain a twicedifferentiable curve. Sim
ilar results can be proved for curves in E 3 using systems of ordinary differential
equations. See Willmore [3].
Sec. 6] SUMMARY 123
(a) k(s) = 1/(1 + s 2 ), (b) k(s) = 1/s (s > 0), where s is the
arc length.
11. Prove that two unitspeed curves a and in E 2 are congruent if and
only if K a = ±K£.
6 Summary
The basic result of this chapter is that an arbitrary isometry of Euclidean
space can be uniquely expressed as an orthogonal transformation followed
by a translation. Its main consequences are that the derivative map of an
isometry F is at every point essentially just the orthogonal part of F, and
that there is a unique isometry which carries any one given frame to
another. Then it is a routine matter to test the concepts introduced earlier
and discover which belong to Euclidean geometry, that is, which are pre
served by isometries of Euclidean space. Finally, we proved an analogue
for curves of the wellknown "sideangleside," "sidesideside" theorems
on triangles from elementary plane geometry. Namely, we showed that
curvature and torsion (and speed) provide a necessary and sufficient con
dition for two given curves to be congruent. Furthermore, the required
isometry can be explicitly computed.
CHAPTER
IV
Calculus on a Surface
This chapter begins with the definition of a surface in E 3 and with some
standard ways to construct surfaces. Although this concept is a moreor
less familiar one, it is not as widely known as it should be that each surface
has a differential and integral calculus strictly comparable with the usual
calculus on the Euclidean plane E 2 . The elements of this calculus — func
tions, vector fields, differential forms, mappings — belong strictly to the
surface and not to the Euclidean space E 3 in which the surface is located.
Indeed, we shall see in the final section that this calculus survives un
damaged when E is removed, leaving just the surface and nothing more.
1 Surfaces in E 3
A surface in E is, to begin with, a subset of E , that is, a certain collection
of points of E 3 . Of course, not all subsets are surfaces: We must certainly
require that a surface be smooth and twodimensional. We shall express
this requirement in mathematical terms by the next two definitions.
1.1 Definition A coordinate patch x: D — > E is a onetoone regular
mapping of an open set D of E 2 into E .
The image x (D ) of a coordinate patch x — that is, the set of all values of
x — is a smooth twodimensional subset of E (Fig. 4.1). Regularity (Defi
nition 7.9 of Chapter I), for a patch as for a curve, is a basic smoothness
condition; the onetoone requirement is included to prevent x(D) from
cutting across itself. Furthermore, in order to avoid certain technical
difficulties (Example 1.7), we shall sometimes use proper patches, those
for which the inverse function x 1 : x(D) — ■> D is continuous (that is, has
124
Sec. 1]
SURFACES IN E 3
125
D
FIG. 4.1
continuous coordinate functions). If we think of D as a thin sheet of rubber,
we can get x (D ) by bending and stretching D in a not too violent fashion.
To construct a suitable definition of surface we start from the rough
idea that any small enough region in a surface M resembles a region in the
plane E 2 . The discussion above shows that this can be stated somewhat
more precisely as: Near each of its points, M can be expressed as the image of
a proper patch. (When the image of a patch x is contained in M, we say
that x is a patch in M. ) To get the final form of the definition, it remains
only to define a neighborhood 91 of p in M to consist of all points of M whose
Euclidean distance from p is less than some number e > 0.
1.2 Definition A surface in E 3 is a subset M of E 3 such that for each point
p of M there exists a proper patch in M whose image contains a neighbor
hood of p in M (Fig. 4.2).
The familiar surfaces used in elementary calculus satisfy this definition;
for example, let us verify that the unit sphere 2 in E 3 is a surface. By defi
nition, 2 consists of all points at unit distance from the origin — that is, all
points p such that
Up 11= (pi 2 + v? + P 3 2 ) 1/2 = i.
To check the definition above, we start by finding a proper patch in 2
covering a neighborhood of the north pole (0, 0, 1). Note that by dropping
FIG. 4.2
126
CALCULUS ON A SURFACE
[Chap. IV
(u, v) ** (u, v, 0)
FIG. 4.3
each point (g x , q 2 , qs) of \the northern hemisphere of 2 onto the xy plane
at (qi, q 2 , 0) we get a onetoone correspondence of this hemisphere with
a disc D of radius 1 in the xy plane (see Fig. 4.3 ) . If we identify this plane
with E 2 by means of the natural association (q lf q 2 , 0) <> (qi, q 2 ), then D
becomes the disc in E 2 consisting of all points (u, v) such that u 2 + v < 1.
Expressing this correspondence as a function on D we find the formula
x(u, v) = (u, v, y/l — v? — v 2 ).
Thus x is a onetoone function from D onto the northern hemisphere
of 2. We claim that x is a proper patch. The coordinate functions of x are
differentiable on D, so x is a mapping. To show that x is regular, we com
pute its Jacobian matrix (or transpose)
fdu <to dp
du du du
du dv df
\dv dv dv
1
1
where / = \/l — u 2 — v 2 . Evidently the rows of this matrix are always
linearly independent, so its rank at each point is 2. Thus by the criterion
following Definition 7.9 of Chapter I, x is regular, and hence is a patch.
Furthermore x is proper, since its inverse function x _1 : x (D ) — > D is given
by the formula
x _1 (pi, P2, pa) = (pi, pa),
hence is certainly continuous. Finally we observe that the patch x covers
a neighborhood of p = (0, 0, 1 ) in S. Indeed it covers a neighborhood of
every point q in the northern hemisphere (Fig. 4.4).
In a strictly analogous way, we can find a proper patch covering each
Sec. 1] SURFACES IN E 3 127
FIG. 4.4
of the other five coordinate hemispheres of 2, and thus verify, by Definition
1.2, that 2 is a surface. Our real purpose here has been to illustrate Defi
nition 1.2 — we shall soon find a much quicker way to prove (in particular)
that spheres are surfaces.
The argument above shows that if / is any differentiate realvalued
function on an open set D in E 2 , then the function x: D — > E such that
x(u, v) = (w, v,f(u, v))
is a proper patch. We shall call patches of this type Monge patches.
We turn now to some standard methods of constructing surfaces. Note
that the image M — x(D) of just one proper patch automatically satisfies
1.2; M is then called a simple surface. (Thus Definition 1.2 says that any
surface in E can be constructed by gluing together simple surfaces.)
1.3 Example The surface M : z = f(x, y). Every differentiable real
valued function / on E 2 determines a surface M in E 3 : the graph of /,
that is, the set of all points of E 3 whose coordinates satisfy the equation
z = f(x, y). Evidently M is the image of the Monge patch
x(u,v) = (u, v,f(u, v));
hence by the remarks above, M is a simple surface.
If g is a realvalued function on E 3 and c is a number, denote by M:
g = c the set of all points p such that g(p) = c. For example, if g is a tem
perature distribution in space, then M: g = c consists of all points of
temperature c. There is a simple condition that tells when such a subset of
E is a surface.
1.4 Theorem Let g be a differentiable realvalued function on E 3 , and c
a number. The subset M: g(x, y, z) = c of E 3 is a surface if the
differential dg is not zero at any point of M.
(In Definition 1.2 and in this theorem we are tacitly assuming that M
has some points in it; thus the equation x 2 + y + z 2 = —1, for example,
does not define a surface. )
128
CALCULUS ON A SURFACE
[Chap. IV
K3>
— ^(Pi , Pi , 0) «» (pi , 1h)
FIG. 4.5
Proof. All we do is give geometric content to a famous result of advanced
calculus — the implicit function theorem. If p is a point of M , we must find
a proper patch covering a neighborhood of p in M (Fig. 4.5 ) . Now
dx dy dz
Thus the hypothesis on dg is equivalent to assuming that at least one of
these partial derivatives is not zero at p, say (dg/dz) (p) ^ 0. In this case,
the implicit function theorem says that nearp the equation g(x, y, z) = c
can be solved for z. More precisely, it asserts that there is a differentiable
realvalued function h defined on a neighborhood D of (pi, pi) such that
(1) For each point (u, v) in D, the point (u, v, h(u, v)) lies in M; that
is, g(u, v, h(u, v)) — c.
(2) Points of the form (u, v, h(u, v)), with (u, v) in D, fill a neighbor
hood of p in M .
It follows immediately that the Monge patch x: D — * E 3 such that
x(w, v) = (u, v,h(u,v))
satisfies the requirements in Definition 1.2. Since p was an arbitrary point
of M , we conclude that M is a surface. 
When M: g = c is a surface, M is said to be defined implicitly by the
equation g — c. It is now very easy to prove that spheres are surfaces. The
sphere S in E 3 of radius r > and center c = (c 1} c 2 , C3) is the set of all
points at distance r from c. If g = 2 (a:* — c») 2 , then S is defined impli
Sec. 1]
SURFACES IN E 3
129
citly by the equation g = r . Now dg =
22 (a;* — Ci)d,Xi, hence dg is zero only at
the point c, which is not in 2. Thus 2 is
a surface.
Using this theorem and the notion of
curve defined on page 20 we derive two
wellknown types of surfaces.
FIG. 4.6
1.5 Example Cylinders. As a line L,
perpendicular to a plane P, moves along
a curve C in P, it sweeps out a cylinder. For definiteness, let P be, say,
the xy plane, so that L is always parallel to the z axis as in Fig. 4.6. If the
curve C is given by
C:f(x,y) = c inE 2 ,
let J be the function on E such that J(pi, p 2 , P3) = f(pi, P2). Then the
resulting cylinder is evidently given by
M:J(x,y,z) = c inE 3 .
The definition of curve on page 20 requires that at each point of C either
df/dx or df/dy is nonzero. Since
(pi,P2,Pa) = %(puP»),
and similarly for d/dy, it follows that dg is never zero at a point of M .
Thus M is a surface.
When C is a circle, we obtain a circular cylinder M : x 2 + y — r 2 in E 3 .
In Example 1.5 we constructed a surface essentially by translating a
curve; now we get one by rotating a curve.
1.6 Example Surfaces of revolution. Let C be a curve in a plane P,
and let A be a line in P which does not meet C. If this profile curve C is
revolved around the axis A, it sweeps out a surface of revolution M in E 3 .
We now check, using Theorem 1.4, that M really is a surface. For sim
plicity, assume that P is a coordinate plane and A is a coordinate axis —
say the xy plane and x axis, respectively. Since C must not meet A, we
assume it is in the upper half (y > 0) of the xy plane. As C is revolved,
each point (#1, q 2 , 0) of C gives rise to a whole circle of points
(qi, #2 cos v, q 2 sin v) in M, for ^ v ^ 2ir.
Put in reverse, a point p = (pi, p 2 , ps) is in M if and only if the point
130
CALCULUS ON A SURFACE
[Chap. IV
P = (Pi, VP2 2 + P3 2 , 0)
is in C (Fig. 4.7).
If the profile cur ve is C: f(x, y) = c, we define a function g on E 3 by
d(x, V, z) = f(x, VV + z 2 ) Then the argument above shows that the re
sulting surface of revolution is exactly M: g(x, y, z) = c. Using the chain
rule, it is not hard to show that dg is never zero on M , so M is a surface.
The circles in M generated, under revolution, by each point of C are
called the parallels of M, the different positions of C as it is rotated are
called the meridians of M. This terminology derives from the geography of
the sphere; however, a sphere is not a surface of revolution as defined above.
Its profile curve must twice meet the axis of revolution, so two "parallels"
reduce to single points. To simplify the statements of subsequent theorems
we use a slightly different terminology in this case; see Exercise 12.
The necessity of the properness condition on the patches in Definition
1.2 is shown by the following example.
1.7 Example Suppose that a rectangular strip of tin is bent into a fig
ure8, as in Fig. 4.8. The configuration M which results does not satisfy
our intuitive picture of what a surface should be, for along the axis A, M
is not like the plane E 2 but is instead like two intersecting planes. To ex
press this construction in mathematical terms, let D be the rectangle
— ir < u < tt, < ^<linE and define x: D — > E 3 by x(u, v) = (sin u,
sin 2u, v). It is easy to check that x is a patch, but its image M = \(D) is
FIG. 4.7
E*
D
FIG. 4.8
Sec. 1]
SURFACES IN E 3
131
?£> *e>"
FIG. 4.9
FIG. 4.10
not a surface: x is not a proper patch. Continuity fails for x _1 : M — > D
since, roughly speaking, to restore M to D, x _1 must tear M along the axis
A (the z axis of E ).
By Example 1.6, the familiar torus of revolution T is a surface (Fig. 4.19).
With somewhat more work, one could construct double toruses of various
shapes, as in Fig. 4.9. By adding "handles" and "tubes" to existing surfaces
one can — in principle, at least — construct surfaces of any desired degree
of complexity (Fig. 4.10).
EXERCISES
1. None of the following subsets M of E 3 are surfaces. At which points p
is it impossible to find a proper patch in M that will cover a neighbor
hood of p in M? (Sketch M — formal proofs not required.)
(a) Cone M: z 2 = x 2 + y 2 .
(b) Closed disc M: x 2 + y 2 ^ 1, z = 0.
(c) Folded plane M : xy = 0, x ^ 0, y ^ 0.
2. A plane in E 3 is a surface M : ax + by + cz = d, where the numbers a,
b, c are necessarily not all zero. Prove that every plane in E 3 may be
described by a vector equation as on page 60.
3. Sketch the general shape of the surface M : z = ax 9 + by 2 in each of
the cases:
132 CALCULUS ON A SURFACE [Chap. IV
(a) a > b > (c) a > b =
(b) a > > b (d) a = b = 0.
4. In which of the following cases is the mapping x: E 2 — > E 3 a patch?
(a) x(u,v) = (v, uv, v). (c) x(w, v) = (u, u 2 , v + y 3 ).
(b) x(w, y) = (w 2 , w 3 , y). (d) x(u, v) = (cos 2iru, sin 2ru, v).
(Rec,all that x is onetoone if and only if x(u, v) = x(u h Vi) implies
(u, v) = (U U Vi).)
5. (a) Prove that M: (x 2 + y 2 f + 3z 2 = 1 is a surface.
(b) For which values of c is M : z(z — 2) + xy = c a surface?
6. Determine the intersection z = of the monkey saddle
M: z = f(x, y), f = y 3  Zyx 2 ,
with the xy plane. On which regions of the plane is/> 0?/ < 0?
How does this surface get its name?
7. Let x: D— >E be a mapping, with
x(u, v) = (x 1 (u,v),x 2 (u,v),xz(u,v)).
(a) Prove that a point p = (pi, P2, Ps) of E 3 is in the image x(D) if
and only if the equations
Pi = Xi(u, v) pi = x 2 (u, v) p 3 = x 3 (u, v)
can be solved for u and v, with (u, v) in D.
(b) If for every point p in x(D) these equations have the unique
solution: u = /i(pi, p 2 , Pz), v = / 2 (pi, Pi, Ps), with (w, v) in D,
prove that x is onetoone and that x 1 : x (D ) ^ D is given by the
formula
x _1 ( P ) = (A(p),/.(p)).
8. Let x: D — > E 3 be the function given by
x(w, v) = (w , W, V )
on the first quadrant D : u > 0, v > 0. Show that x is onetoone and
find a formula for its inverse function x _1 : x(D) — > D. Then prove
that x is a proper patch.
9. Let x: E 2 — > E 3 be the mapping
x(u, v) = {u { v, u — v, uv).
Show that x is a proper patch, and that the image of x is the surface
M.z = (x 2  y 2 )/4:.
10. If F is an isometry of E 3 and M is a surface in E 3 , prove that the image
Sec. 2] PATCH COMPUTATIONS 133
FIG. 4.11
F(M) is also a surface in E 3 . (Hint: If x is a patch in M, then the
composite function F(x) is regular, since F(x)* = F* x* by Ex. 12
of 1.7.)
11. The assertion in Exercise 10 remains true when F is merely a diffeo
morphism. Prove this special case: If F is a diffeomorphism of E 3 , then
the image of the surface M: g = c is M: g = c, where g = g(F~ ) —
and M is a surface. (Hint: If dg(v) 9* at pin M , show by using Ex. 9
of 1.7 that dg (F* v) 5* 0.
12. If / is a differentiable function and f(x,y 2 ) = c defines a curve C in
the xy plane, then C is symmetric about the x axis and must cross this
axis once (if C is an arc) or twice (if C is closed). Prove that revolving
C about the x axis gives a surface M in E 3 . We call M an augmented
surface of revolution: If the points on the axis are deleted, it becomes
an ordinary surface of revolution (Fig. 4.11).
2 Patch Computations
In Section 1, coordinate patches were used to define a surface; now we con
sider some properties of patches that will be useful in studying surfaces.
Let x: D — » E 3 be a coordinate patch. Holding u or v constant in the
function (w, v) — * x(w, v) produces curves. Explicitly, for each point
(uo, vo) in D the curve
u —*■ x(u, Vo)
is called the uparameter curve, v = Vo, of x; and the curve
v — > x(u , v)
is the vparameter curve, u = u (Fig. 4.12).
Thus the image x(D) is covered by these two families of curves, which
are the images under x of the horizontal and vertical lines in D, and one
curve from each family goes through each point of x(D).
134
CALCULUS ON A SURFACE
[Chap. IV
V
(Wo , V )
upaxameter curves
u = Mo.
D
( Mparameter
curves
E a
u
x(mo , v )
V = v
FIG. 4.12
x«(mo , «o)
x.(wo , t> )
2.1 Definition If x: D > E 3 is
x(t*o , i>o)
FIG. 4.13
a patch, for each point (wo, v ) in
D:
(1) The velocity vector at Uq
of the wparameter curve, v = v ,
is denoted by x u (uo, v ).
(2) The velocity vector at y
of the yparameter curve, u = Uo,
is denoted by x v (u , v ).
The vectors x u (uo, v ) and x v (u , v ) are called the partial velocities of x
at (uo,v ) (Fig. 4.13).
Thus x„ and x„ are actually functions on D whose values at each point
(u , v ) are tangent vectors to E at x(u , v ). The subscripts u and v are
intended to suggest partial differentiation. Indeed if the patch is given in
terms of its Euclidean coordinate functions by a formula
x(u,v) = (xi(u,v),x 2 (u,v),X3(u,v)),
then it follows from the definition above that the partial velocity functions
are given by
(dx\ dx 2 dxz\
du ' du du Jx
(dxi dXi dxs\
dv ' dv dv/x
The subscript x (frequently omitted ) is a reminder that x u (u, v) and x„ (u, v)
have point of application x(u, v).
2.2 Example The geographical patch in the sphere. Let 2 be the sphere
of radius r > centered at the origin of E . Longitude and latitude on the
earth suggest a patch in S quite different from the Monge patch used on
Sec. 2]
PATCH COMPUTATIONS
135
D
<u,v)
FIG. 4.U
S in Section 1. The point x(u, v) of 2 with longitude u ( — k < u < ir)
and latitude v ( — x/2 < v < ir/ 2 ) has Euclidean coordinates (Fig. 4.14).
x(u, v) = (r cos v cos w, r cos y sin w, r sin v).
With the domain D of x denned by these inequalities, the image x(D)
of x is all of 2 except one semicircle from north pole to south pole. The
wparameter curve, v = vo, is a circle — the parallel of latitude vo. The
vparameter curve, u = u , is a semicircle — the meridian of longitude uo.
We compute the partial velocities of x to be
x u (u, v) = r( — cos v sin u, cos v cos u, 0)
x^w, v) = r( — sin v cos u, —sin v sin w, cos i>)
where r denotes a scalar multiplication. Evidently x u always points due
east, and x„ due north. In a moment we shall give a formal proof that x is
a patch in 2 (Fig. 4.15).
To test whether a given subset M of E 3 is a surface, Definition 1.2 de
mands proper patches (and Example 1.7 shows why). But once we know
that M is a surface, the properness condition need no longer concern us
(Exercise 14 of Section 3). Furthermore, in many situations the onetoone
restriction on patches can also be dropped.
2.3 Definition A regular mapping x: D — » E 3 whose image lies in a surface
M is called a parametrization of the region x(Z>) in M.
(Thus a patch is merely a onetoone parametrization.) In favorable
cases this image x(D) may be the whole surface M, and we have then the
analogue of the more familiar notion of parametrization of a curve (p. 20).
Parametrizations will be of first importance in practical computations with
136
CALCULUS ON A SURFACE
[Chap. IV
x„(Wo , »o)
meridian
u — Mo
FIG. 4.15
surfaces, so we consider some ways of determining whether a mapping
x: D — ► E is a parametrization of (part of ) a given surface M.
The image of x must, of course, lie in M. Note that if the surface is given
in the implicit form M: g = c, this means that the composite function g(x)
must have constant value c.
To test whether x is regular, note first that parameter curves and partial
velocities x„ and x„ are welldefined for an arbitrary differentiable mapping
x: D — > E . Now the last two rows of the cross product
*7i U* U s
dxi
x u X x„ = du
dxi
~dv
give the (transposed) Jacobian matrix of x at each point. Thus the regu
larity of x is equivalent to the condition that x M X x„ is never zero, or, by
properties of the cross product, that at each point (u, v) of D the partial ve
locity vectors of x are linearly independent.
Let us try out these methods on the mapping x given in Example 2.2.
Since the sphere is defined implicitly by g = x 2 + y 2 + z 2 = r 2 , we must
show that g (x) = r 2 . Substituting the coordinate functions of x for x, y, and
z, we get
r~ 2 g(x) = (cos v cos u) 2 f (cos v sin u) 2 + sin 2 u
= cos 2 y + sin 2 ?; = 1.
A short computation using the formulas for x u and x„ given in Example
2.2 yields
r~ 2 x„ X x„ = cos u cos 2 y U\ + sin u cos 2 y £/ 2 + cos v sin v U 3 .
dx 2
du
dx 3
du
dx%
dv
dxz
dv
See. 2]
PATCH COMPUTATIONS
137
Since tt/2 < v < x/2 for the domain D of x, cos v is never zero there;
but sin u and cos u are never simultaneously zero, so x u X x„ is never zero
on D. Thus x is regular, and hence is a parameterization.
To show that x is a patch, we prove it is onetoone, that is, show that
x(w, v) = x(wi, vi) implies (u, v) = (wi, Vi). If x(w, y) = x(ui, Vi), then
the definition of x gives three coordinate equations:
r cos v cos u — r cos Wi cos U\
r cos v sin u = r cos Vi sin Wi
r sin v = r sin w x .
Again, since w/2 < v < tt/2 for all points of D, the last equation
implies that v = Vi. Thus r cos v = r cos t>i > may be cancelled from the
first two equations and we conclude that u = U\ also.
For this particular function x in 2, these results might almost be con
sidered obvious from the discussion in Example 2.2, but the methods used
above will serve in more difficult cases.
We shall now see how to find natural parametrizations in cylinders and
surfaces of revolution.
2.4 Example Parametrization of a cylinder M. Suppose, as in Ex
ample 1.5, that M is the cylinder over a curve C: f(z, y) = a in the xy
plane (Fig. 4.16). If a = («i, a 2 , 0) is a parametrization of C, we assert
that
x(u,v) = («i(w), a 2 (u),v)
(ai(u), ai(w), 0)
FIG. 4.16
138
CALCULUS ON A SURFACE
[Chap. IV
is a parametrization of M . Clearly x lies in M, covers all of M , and is differ
entiable. Furthermore x is regular, for at each point (u, v) the partial
velocities
_ /dai daz \
Xu " \du ' du ' V
x, = (0, 0, 1)
are linearly independent (x u is never zero, since a is by definition regular).
If the curve a is defined on an interval J, the domain of x is the vertical
strip D: u in /, v arbitrary. (Thus if / is the whole real line, D is E 2 .) The
wparameter curves of x are merely translates of C and are called the
crosssectional curves of the cylinder. The vparameter curves follow the
straight lines called the rulings (or "elements") of the cylinder. If C is not
a closed curve, then a — and hence also x— is onetoone, so x is a patch.
But if C is closed, x wraps D an infinite number of times around C.
2.5 Example Parametrization of a surface of revolution. Suppose that
M is obtained, as in Example 1.6, by revolving a curve C in the upper half
of the xy plane about the x axis. Now let
a(u) = (g(u),h(u), 0)
be a parametrization of C (note that h > 0). As we observed in Example
1.6, when the point (g(u), h(u), 0) on the profile curve C has been rotated
through an angle v, it reaches a point x(u, v) with the same x coordinate
g(u), but new y and z coordinates h(u) cos v and h(u) sin v, respectively
(Fig. 4.17). Thus
x(u, v) = (g(u), h(u) cos v, h(u) sin v)
(jf(u), *(tt), 0)
Z
FIG. 4.17
Sec. 2]
PATCH COMPUTATIONS
139
Evidently this formula defines a mapping into M whose image is all of M.
A short computation shows that x„ and x v are always linearly independent,
so x is a parametrization of M. As in Example 2.4, the domain D of x con
sists of all points (u, v) for which u is in the domain of a. The wparameter
curves of x parametrize the meridians of M; the vparameter curves, the
parallels. (Thus the parametrization x: D —*■ M is never onetoone.)
Obviously we are not limited to rotating curves in the xy plane about
the x axis. But with other choices of coordinates, we maintain the same
geometric meaning for the functions g and h: g measures distance along
the axis of revolution, while h measures distance from the axis of revolu
tion.
Actually the geographical patch in the sphere is one instance of Example
2.5 (with u and v reversed) ; here is another.
2.6 Example Torus of revolution T. This is the surface of revolution
obtained when the profile curve C is a circle. Suppose that C is the circle
in the xz plane with radius r > and center (R, 0, 0). We shall rotate
about the z axis; hence we must require R > r to keep C from meeting
the axis of revolution. A natural parametrization (Fig. 4.18) for C is
a(u) = (R \ r cos u, r sin u).
Thus by the remarks above we must have g(u) = r sin u, distance along
the z axis, and h(u) = R + r cos u, distance from the z axis. The general
argument in Example 2.5 — with coordinate axes permuted — then yields the
parametrization
x(u, v) = (h(u) cos v, h(u) sin v, g{u))
= ((R \ r cos u) cos v, (R + r cos u) sin v, r sin u).
The domain of x is the whole plane E 2 , and (as always when the profile
curve is closed) x is periodic in both u and v. Here
x(m + 2ir, v + 2x) = x(w, v) for all (u, v).
There are infinitely many different parametrizations (and patches) in
FIG 4.18
FIG. 4.19
140 CALCULUS ON A SURFACE [Chap. IV
any surface. Those we have discussed have been singled out by the natural
way they are fitted to their surfaces.
EXERCISES
1. Find a parametrization for the entire surface obtained by revolving:
(a) C: y = cosh x around the x axis (catenoid).
(b) C: (z — 2) 2 + y 2 = 1 around the y axis (torus).
(c) C: z = x 2 around the z axis (paraboloid of revolution).
2. Partial velocities x M and x„ are defined for an arbitrary mapping x:
D — > E , so we may consider the realvalued functions
E = x u »x u F = x u «x„ G = x v *x v
on D. Prove
 x u X x v  2 = EG F 2 .
Deduce that x is a regular mapping if and only if EG — F 2 is never
zero. (This is often the easiest way to check regularity. The geometric
significance of these functions is discussed in V.4).
3. Show that
M: (y/ x * + y 2  4) 2 + z = 4
is a torus of revolution: Find a profile circle and the axis of revolution.
A ruled surface is a surface swept out by a straight line L moving
along a curve 0. The various positions of the generating line L are
called the rulings of the surface. Such a surface thus always has a
parametrization in ruled form
x(u,v)=p(u) + v8(u) or @(v) + u8(v)
where we call /J the base curve, 8 the director curve. Alternatively we
may vizualize 8 as a vector field on /?. Frequently it is necessary to
restrict v to some interval, so the rulings may not be entire straight
lines.
4. Show that the saddle surface M: z = xy is doubly ruled: Find two
ruled parametrizations with different rulings.
5. A cone is a ruled surface with parametrization of the form
x(u, v) = p + v 8(u).
Thus all rulings pass through the vertex p (Fig. 4.20). Show that the
regularity of x is equivalent to both v and 5X5' never zero. (Thus the
vertex is never part of the cone. )
Sec. 2]
PATCH COMPUTATIONS
141
FIG. 4.20
FIG. 4.21
6. A cylinder is a ruled surface with parametrization of the form
x(u, v) = 8(w) + vq.
Thus the rulings are all parallel (Fig. 4.21). Prove that the regularity
of x is equivalent to £ X q never zero. Show that this definition
generalizes Example 2.4.
7. A line L is attached orthogonally to an axis A (Fig. 4.22 ). If L moves
along A and rotates — both at constant speed — then L sweeps out a
helicoid H.
If A is the z axis, then H is the image of the mapping x: E 2 — > E 3 such
that
x(u, v) = (u cos v, u sin v, bv)
(b * 0).
(a) Prove that x is a patch.
(b ) Describe the parameter curves
of x.
(c ) Express the helicoid in implicit
form g = c.
8. (a) Show that x: D ,— > E 3 is a reg
ular mapping, where
\(u, v) = (u cos v, u, sin v)
onD:u>0. /^ (m cos v, u sin t>, 0)
(b) Find a function g(x, y, z) such
that the image of x is the sur fig. 4.22
face M : g = 0.
(c) Show that M is a ruled surface and sketch M. (Hint: Begin with
the curve sliced from M by the plane y = 1.)
9. Let jS be a unitspeed parametrization of the unit circle in the xy plane.
(0,0,
142 CALCULUS ON A SURFACE [Chap. IV
Construct a ruled surface as follows: Move a line L along in such a
way that L is always orthogonal to the radius of the circle and makes
constant angle 7r/4 with &' (Fig. 4.23).
(a) Derive this parametrization of the resulting ruled surface M:
x(u,v) = 0(u) + v(p'(u) + Uz).
(b) Express x explicitly in terms of v
and coordinate functions for 0.
(c) Deduce that M is given implicitly
by the equation
x 2 + y 2  i = 1.
(d) Show that if the angle 7r/4 above
is changed to — t/4, the same surface M results. Thus M is
doubly ruled.
(e) Sketch this surface M showing the two rulings through each of
the points (1, 0, 0) and (2, 1, 2).
A quadrie surface is a surface M: g = for which g involves at most
quadratic terms in x x , x 2 , x 3 — that is,
9 = iLi anXiXj + 2 Mi + c.
i,j i
Trivial cases excepted, there are just five types of quadrie surfaces, repre
sented by the familiar surfaces in the next three examples (see Theorem
2.2, p. 280, Birkhoff and MacLane [2]).
10. In each case, (i) prove that M is a surface and sketch its general
shape, (ii) show that x is a parametrization and find its image in M.
(a) Ellipsoid, M: 2 + l + Z  = l
a 2 b 2 c 2
x(u, v) = (a cos u cos v, b cos u sin v, c sin u)
on/): x/2 < u < x/2.
2 2 2
X 1J z
(b) Elliptic hyperboloid, M: — + ^ = 1
a 2 b 2 c 2
x(u, v) = (a cosh u cos v, b cosh u sin v, c sinh u) on E 2 .
2 2 2
(c) Elliptic hyperboloid (two sheets), M: ( — = —1
a 2 b 2 c 2
x (u, v) = (a sinh u cos v, b sinh u sin v, c cosh u) onD.u^O.
2 2
X 1J
11. Elliptic paraboloid, M: z = — + f .
a 2 o 2
(a) Show that M is a surface, and that
Sec. 3] DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS 143
x(u, v) = (au cos v, bu sin v, w 2 ), u > 0,
is a parametrization that omits only one point of M.
(b) Describe the parameter curves of x in general, and sketch this
surface for a = 1, b = 4, showing some parameter curves.
2 2
X V
12. Hyperbolic paraboloid, M: z =   — fr .
a 2 cr
(a) Show that x: E 2 — > E is a proper patch covering all of M, where
x(u, v) = (a(u f v), b(u — v), 4uv).
(b) Show that M is a doubly ruled surface by rewriting x in ruled
form in two different ways.
(c) Same as (b) of Exercise 11.
13. Let M be the surface of revolution obtained by revolving the curve
t — > (g(t), h(t), 0) about the x axis (h > 0). Show that
(a) If gr is never zero, M has a parametrization of form
x(w, v) = (u, f(u) cos v, f(u) sin v).
(b) If h is never zero, M has a parametrization of form
x(w, v) = (f(u), u cos v, u sin v).
3 Differentia ble Functions and Tangent Vectors
We now begin an exposition of the calculus on a surface M in E 3 . The
space E will gradually fade out of the picture, since our ultimate goal is a
calculus for M alone. Generally speaking, we shall follow the order of topics
in Chapter I, making such changes as are necessary to adapt the calculus
of the plane E 2 to a surface M.
Suppose that / is a realvalued function denned only on a surface M.
If x: D — > M is a coordinate patch in M, then the composite function
/(x) is called a coordinate expression for /; it is an ordinary real valued
function (u,v) — > /(x (u, v) ). We define / to be differentiable provided all its
coordinate expressions are differentiable in the usual Euclidean sense
(Definition 1.3 of Chapter I).
For a function F: E n — > M, each patch x in M gives a coordinate expres
sion x~ (F) for F. Evidently this composite function is defined only on the
set G of all points p of E n such that F(p) is in x(D). Again we define F
to be differentiable provided all its coordinate expressions are differentiable
in the usual Euclidean sense. [We must understand that this includes the
144
CALCULUS ON A SURFACE
[Chap. IV
requirement that e be an open set of E n , so that the differentiability of
x 1 (F) : — >• E 2 is welldefined, as in Section 7 of Chapter I.]
In particular, a curve a: I — > M in a surface M is, as before, a differentia
ble function from an open interval / into M.
To see how this definition works out in practice, we examine an important
special case.
3.1 Lemma If a is a curve a: I —*■ M whose route lies in the image x(D)
of a single patch x, then there exist unique differentiate functions Oi, ch
on / such that
a(t) = x(ai(t), a 2 (t)) for all I
or, in functional notation, a = x(a u a 2 ). (See Fig. 4.24.)
Proof. By definition, the coordinate expression x~ x a: J — *■ D is differentia
ble — it is just a curve in E 2 whose route lies in the domain D of x. If a h az
are the Euclidean coordinate functions of x 1 a, then
a = xx a = x(oi, 02).
These are the only such functions, for if a = x(6i, 6 2 ), then
(«i, 02) = x _1 a = x _1 x(6i, 6 2 ) = (61, &2>. I
These functions ai, a 2 are called the coordinate functions of the curve a
with respect to the patch x. For example, the curve a given in (3 ) of Exam
ple 4.2, Chapter I, lies in the part of the sphere 2 of radius 2 that is covered
by the patch x given in Example 2.2. Observe that this curve moves so as
to have equal longitude and latitude at each point. In fact its coordinate
functions with respect to x are ai(t) = (h(t) = t, since by the formula
for x,
x(a,i(t), (h(t)) = x(t, t) = 2(cos 2 f, cos t sin t, sin t) = a(t).
For an arbitrary patch x: D — > M (as in the case just considered) it is
<x(t)= x( ai (t), 0,(0)
FIG. 4.24
Sec. 3]
DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS
145
natural to think of the domain D as a map of the region x(D) in M. The
functions x and x _1 establish a onetoone correspondence between objects
in x(Z>) and objects in D. If a curve a in x(D) represents the voyage of a
ship, the coordinate curve (ai, ck) plots its position on the map D.
A rigorous proof of the following rather technical fact requires the meth
ods of advanced calculus, and we shall not attempt to give a proof here.
3.2 Theorem Let M be a surface in E 3 . If F : E n — > E 3 is a (diff erentiable )
mapping whose image lies in M, then considered as a function F: E n — ► M
into M, F is differentiable (as on p. 143 . )
This theorem links the calculus of M tightly to the calculus of E 3 . For
example, it implies the "obvious" result that a curve in E 3 which lies in
Mis a curve of M.
Since a patch is a differentiable function from (an open set of) E into
E 3 , it follows that a patch is a differentiable function into M . Hence its
coordinate expressions are all differentiable, so patches overlap smoothly.
3.3 Corollary If x and y are patches in a surface M in E whose images
overlap, then the composite functions x _1 y and y _1 x are (differentiable)
mappings defined on open sets of E 2 .
(The function y 1 x, for example, is defined only for those points (u, v)
in D such that x(u,v) lies in the image y(E) of y (Fig. 4.25).
By an argument like that for Lemma 3.1, Corollary 3.3 can be rewritten.
3.4 Corollary If x and y are overlapping patches in M, then there exist
unique differentiable functions u and v such that
y(u,v) = x(u(u,v),v(u,v))
for all (u, v) in the domain of x 1 y. In functional notation: y = x(w, v).
There are, of course, symmetrical equations expressing x in terms of y.
Corollary 3.3 makes it much easier to prove differentiability. For exam
ple, if / is a real valued function on M, instead of verifying that all coordi
146 CALCULUS ON A SURFACE [Chap. IV
nate expressions /(x) are Euclidean differentiable, we need only do so for
enough patches x to cover all of M (so a single patch will often be enough).
The proof is an exercise in checking domains of composite functions:
For an arbitrary patch y, fx and x _1 y differentiable imply /xx 1 y differen
tiable. This function is in general not /y, because its domain is too small.
But since there are enough x's to cover M , such functions constitute all
of /y, and thus prove it is differentiable.
It is intuitively clear what it means for a vector to be tangent to a
surface M in E 3 . A formal definition can be based on the idea that a curve in
M must have all its velocity vectors tangent to M.
3.5 Definition Let p be a point of a surface M in E 3 . A tangent vector
v to E at p is tangent to M at p provided v is a velocity vector of some
curve in M (Fig. 4.26).
The set of all tangent vectors to M at p is called the tangent plane of M
at p, and is denoted by T P (M). The following result shows, in particular,
that at each point p of M the tangent plane T p (M) is actually a twodimen
sional vector subspace of the tangent space T P (E 3 ).
3.6 Lemma Let p be a point of a surface M in E 3 , and let x be a patch
in M such that x(w , v ) = p. A tangent vector v to E 3 at p is tangent
to M if and only if v can be written as a linear combination of x u (u Q , v )
and x v (u , v ).
Since partial velocities are always linearly independent, we deduce that
they provided a basis for the tangent plane of M at each point of x(D).
Proof. Note that the parameter curves of x are curves in M, so these
partial velocities are tangent to M at p.
First suppose that v is tangent to M at p; thus there is a curve a in M
such that a(0) = p and a (0) = v. Now by Lemma 3.1, a may be written
a = x(ai, a 2 ); hence by the chain rule
T V (M)
FIG. 4.26
Sec. 3] DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS 147
/ / \ da\ , N da,2
a = x„Ui, a 2 J j + x„(ai, a 2 ; = •
at at
But since a(0) = p = x(w , v ), we have Oi(0) = u , a 2 (0) = i>o. Hence
evaluation at t = yields
v = a'(0) = r 1 (0) x„(w , Wo) + rr (0) x "( w o, Wo).
at at
Conversely, suppose that a tangent vector v to E 3 can be written
v = Cix u (u , v ) + c 2 x„(w , v ).
By computations as above, v is the velocity vector at t — of the curve
t — > x(w + tci, v Q + <c 2 ).
Thus v is tangent to M at p. 
A reasonable deduction, based on the general properties of derivatives,
is that the tangent plane T P {M) is the linear approximation of the surface
M near p.
3.7 Definition A Euclidean vector field Z on a surface M in E 3 is a func
tion that assigns to each point p of M a tangent vector Z(p) to E 3 at p.
A Euclidean vector field V for which each vector V(p) is tangent to M
atp is called a tangent vector field on M (Fig. 4.27). Frequently these vector
fields are defined, not on all of M, but only on some region in M . As usual,
we always assume differentiability (Exercise 12).
A Euclidean vector z at a point p of M is normal to M if it is orthogonal
to the tangent plane T P {M) — that is, to every tangent vector to M at p.
And a Euclidean vector field Z on M is a normal vector field on M provided
each vector Z(p) is normal to M.
Because T P (M) is a twodimensional subspace of ^(E ), there is only
one direction normal to M at p: All normal vectors z at p are collinear.
Z(p)>
FIG. 4.27
148 CALCULUS ON A SURFACE [Chap. IV
Thus if z is not zero, it follows that T P (M) consists of precisely those vectors
in T P (F, ) that are orthogonal to z.
It is particularly easy to deal with tangent and normal vector fields on a
surface given in implicit form.
3.8 Lemma If M: g = c is a surface in E 3 , then the gradient vector field
Vg = X (dg/dXi)Ui (considered only at points of M ) is a non vanishing
normal vector field on the entire surface M .
Proof. The gradient is nonvanishing (that is, never zero) on M since by
Theorem 1.4 the partial derivatives dg/dxi cannot simultaneously be
zero at any point of M.
We must show that (V</)(p)»v = for every tangent vector v to M
at p. First note that if a is a curve in M, then g(a) = g(ai, a 2 , 0:3) has
constant value c. Thus by the chain rule
dXi at
Now choose a to have initial velocity
a'(0) = v = (v u V 2 , V 3 )
ata(O) = p. Then
°" Eg wo)) f (0)  £g(p)* " (V9)(pW " a ■
3.9 Example Vector fields on the sphere 2 : g = 2x, 2 = r 2 . The lemma
shows that
X = \ Vg = S XiUi
is a normal vector field on S (Fig. 4.28). This is geometrically evident, since
X(p) = 2 PiUi(p) is the vector p with point of application p! It follows
by a remark above that \ p is tangent to 2 if and only if the dot product
*(P) = Pp
FIG. 4.28
Sec. 3] DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS 149
y p'Pp — v«p is zero. Similarly a vector field V on 2 is a tangent vector
field if and only if V'X = 0. For example, F(p) = (— p 2 , pi, 0) defines
a tangent vector field on 2 that points "due east" and vanishes at the north
and south poles (0, 0, ±r).
We must emphasize that only the tangent vector fields on M belong to
the calculus of M itself, since they derive ultimately from curves in M
(Definition 3.5). This is certainly not the case with normal vector fields.
However, as we shall see in the next chapter, normal vector fields are quite
useful in studying M from the viewpoint of an observer in E 3 .
Finally, we shall adapt the notion of directional derivative to a surface.
Definition 3.1 of Chapter I uses straight lines in E ; thus we must use the
less restrictive formulation based on Lemma 4.6 of Chapter I.
3.10 Definition Let v be a tangent vector to M at p, and let / be a differ
entiate realvalued function on M. The derivative v[/] of f with respect to v
is the common value of (d/dt)(fa) (0) for all curves a in M which have
initial velocity v.
Directional derivatives on a surface have exactly the same linear and
Leibnizian properties as in the Euclidean case (Theorem 3.3 of Chapter I) .
EXERCISES
1. Let x be the geographical patch in the sphere 2 (Ex. 2.2). Find the
coordinate expression /(x) for the following functions on 2:
(a) /(p) = Pl 2 + p 2 2 , (b) f(p) = ( Pl  p 2 ) 2 + Pz
2. Let x be the parametrization of the torus given in Example 2.6.
(a) Find the Euclidean coordinates ai, a 2 , a 3 of the curve a (t) = x (t, t).
(b) Show that a is periodic, and find its period (see p. 20).
3. (a) Prove Corollary 3.4.
(b) Derive the "chain rule"
_ du dv _ du dv
Yu ~ du Xu + ^ Xv Yv ~ dv~ Xu + dv Xv
where x„ and x„ are evaluated on (u,v).
(c) Deduce that y„ X y, = Jx u X x„, where J is the Jacobian of the
mapping x 1 y = (u,v): D — > E 2 .
4. Let x be a patch in M .
(a) If x* is the derivative map of x (1.7), show that
x*(£/i) = x u x*(£/ 2 ) = x„
where Ui, U 2 is the natural frame field on E 2 .
150 CALCULUS ON A SURFACE [Chap. IV
(b) If / is a differentiable function on M , prove
xJ/] = ^(/(x)) x v [f] =!;(/(*)).
5. Prove that :
(a) v is tangent to M: z  f(x, y) at a point p of M if and only if
Vz = dx ^ Ph v ^ Vl + d ^ Ph V ^ V2 '
(b) if x is a patch in an arbitrary surface M, then v is tangent to M
at x(u, v) if and only if
vx„(w, v) X x„(i*, v) = 0.
6. Let x and y be the patches in the unit sphere 2 that are defined on the
unit disc D: u 2 + v < 1 by
x(u,v) = (u,v,f(u,v)) y(u,v) = (v,f(u,v),u)
where/ = \/l — u 2 — v 2 .
(a) On a sketch of 2 indicate the images x(D) and y(D), and the
region on which they overlap.
(b) At which points of D is y 1 x defined? Find a formula for this
function.
(c) At which points of D is x _1 y defined? Find a formula for this
function.
7. Find a nonvanishing normal vector field onM: z = xy and two tangent
vector fields that are linearly independent at each point.
8. Let C be the right circular cone parametrized by
x(u, v) = y(cos u, sin u, 1).
If a is the curve a(t) = x(\/2t, e l )
(a) Express a in terms of x„ and x„.
(b) Show that at each point of a, the velocity a bisects the angle
between x M and x„. {Hint: Verify that a *x u /\\ x u  = a •x„/ x„ ,
where Xu and x» are evaluated on (y/2t, e 1 ).)
(c) Make a sketch of the cone C showing the curve a.
9. If z is a nonzero vector normal to M at p, let T P (M) be the plane
through p orthogonal to z (see page 60). Prove:
(a) If each tangent vector \ p to M at p is replaced by its tip p + v,
then T P (M) becomes T P (M). (Thus f p {M) gives a concrete
representation of T p (M ) in E 3 . )
(b) If x is a patch in M, then f\( U ,v)(M) consists of all points r in E
such that (r — x(u,v))*x u (u, v) X x v (u, v) = 0.
Sec. 3] DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS 151
(c) If M is given implicitly by g = c, then f p (M) consists of all
points r in E 3 such that (r — p) • (Vg) (p) = 0.
10. In each case find an equation of the form ax + by + cz = d for the
plane T P (M).
(a) p = (0, 0, 0) and M is the sphere
x 2 + y 2 + (z 1) 2 = 1.
(b) p = (1, —2, 3) and M is the ellipsoid
2 2 2
4 ^ 16 ^ 18
(c) p = x(2, 7r/4), where M is the helicoid parametrized by
x(u, v) = (u cos v, u sin v, 2v).
1 1 . (Continuation of Exercise 2 ) .
(a) If m and n are integers with greatest common divisor 1, show that
a(t) = x(mt, nt) is a simple closed curve on the torus, and find
its period. f
(b) If q is an irrational number, show that the curve a: R — >• T such
that <x(t) = x(t, qt) is onetoone.
This curve, called a winding line on the torus T, is dense in
T; that is, given any number e > 0, a comes within distance e
of every point of T.
12. A Euclidean vector field Z = J] z i Ui on M is differentiable provided
its coordinate functions z u z 2 , z 3 (on M) are differentiable. If V is a
tangent vector field, show that
(a) For every patch x: D — > M, V can be written as
V(x(u,v)) = f(u,v)x u (u,v) + g(u,v)x v (u,v)
(b) V is differentiable if and only if the functions / and g (on D) are
differentiable.
The following exercises require some knowledge of pointset topology.
They deal with open sets in a surface M in E 3 , that is, sets ai in M which
contain a neighborhood in M of each of their points.
13. Prove that if y: E — > M is a proper patch, then y carries open sets in
E to open sets in M. Deduce that if x: D > M is an arbitrary patch,
then the image x(D) is an open set in M. (Hint: To prove the latter
assertion, use Corollary 3.3.)
t That is, show a(t') = a(t) if and only if t' — t is a multiple of the period p.
Roughly speaking, this means the route of a is Oshaped rather than, say, 8shaped.
152 CALCULUS ON A SURFACE [Chap. IV
14. Prove that every patch x: D — » M in a surface M in E 3 is proper.
(Hint: Use Exercise 13. Note that (x _1 y)y _1 is continuous and agrees
with x _1 on an open set in x(D).)
15. If 01 is a subset of a surface M in E 3 , prove that 01 is itself a surface
in E if and only if 01 is an open set of M .
4 Differential Forms on a Surface
In Chapter I we discussed differential forms on E 3 only in sufficient detail
to take care of the Cartan structural equations (Theorem 8.3 of Chapter
II). In the next three sections we shall give a rather complete treatment of
forms on a surface.
Forms are just what we need to describe the geometry of a surface
(Chapters VI and VII), but this is only one example of their usefulness.
Surfaces and Euclidean spaces are merely special cases of the general notion
of manifold (Section 8). Every manifold has a differential and integral
calculus — expressed in terms of forms — which generalizes the usual ele
mentary calculus on the real line. Thus forms are fundamental to all the
many branches of mathematics and its applications that are based on cal
culus. In the special case of a surface, the calculus of forms is rather easy,
but still gives a remarkably accurate picture of the most general case.
Just as for E 3 , a 0formf on a surface M is simply a (differentiable) real
valued function on M, and a 1form <f> on M is a real valued function on
tangent vectors to M that is linear at each point (Definition 5.1 of Chapter
I). We did not give a precise definition of 2forms in Chapter I, but we shall
do so now. A 2formwillbe a twodimensional analogue of a 1form: a real
valued function, not on single tangent vectors, but on pairs of tangent
vectors. (In this context the term "pair" will always imply that the tangent
vectors have the same point of application.)
4.1 Definition A 2form y\ on a surface M is a realvalued function on all
ordered pairs of tangent vectors v,w to I such that
(1 ) t] (v, w) is linear in v and in w.
(2) r/(v,w) = t)(w,v).
Since a surface is twodimensional, all p forms with p > 2 are zero, by
definition. This fact considerably simplifies the theory of differential forms
on a surface.
At the end of this section we will show that our new definitions are con
sistent with the informal exposition given in Chapter I, Section 6.
Forms are added in the usual pointwise fashion; we add only forms of
the same degree p = 0, 1, 2. Just as we evaluated a 1form <j> on a vector
Sec. 4] DIFFERENTIAL FORMS ON A SURFACE 153
field V, we now evaluate a 2form ijona pair of vector fields V, W to get a
real valued function ri(V,W) on the surface M. Of course we shall always
assume that the forms we deal with are differentiable — that is, convert
(differentiable) vector fields into differentiable functions.
Note that the alternation rule (2) in Definition 4.1 implies that
v(v,\) =
for any tangent vector v. This rule also shows that 2forms are related to
determinants.
4.2 Lemma Let if be a 2form on a surface M, and let v and w be (line
arly independent) tangent vectors at some point of M. Then
?/(av + bw, cv + dw) =
a b
c d
v(y, w)
Proof. Since 17 is linear in its first variable, its value on the pair of tangent
vectors av + bw, cv + dw is av(\, cv + dw) + brj (w, cv + dw). Using
the linearity of i\ in its second variable, we get
ac t?(v,v) + adrj(\,w) + be r?(w,v) + bdrf(w,w).
Then the alternation rule (2 ) gives
77 (av + bw, cv + dw) = (ad — be) ??(v,w). 
Thus the values of a 2form on all pairs of tangent vectors at a point
are completely determined by its value on any one linearly independent pair.
This remark is used frequently in later work.
Wherever they appear, differential forms satisfy certain general proper
ties, established (at least partially) in Chapter I for forms on E 3 . To begin
with, the wedge product of a pform and a qform is always a (p + q)form.
If p or q is zero, the wedge product is just the usual multiplication by a
function. On a surface, the wedge product is always zero if p + q > 2. So
we need a definition only for the case p = q = 1.
4.3 Definition If <f> and \p are 1forms on a surface M, the wedge product
4> a \p is the 2form on M such that
(<t> a ^)(v, w) = 0(v) ^(w)  <f>(w) rf/(\)
for all pairs v, w of tangent vectors to M.
Note that <f> a if/ really is a 2form on M, since it is a real valued function
on all pairs of tangent vectors and satisfies the conditions in Definition 4.1.
The wedge product has all the usual algebraic properties except commuta
tivity; in general, if £ is a pform and 17 is a qform, then
€ A „ = (!)%, A $.
154 CALCULUS ON A SURFACE [Chap. IV
On a surface the only minus sign occurs in the multiplication of 1 forms,
where just as in Chapter I, we have <£ a \f/ = — \f> a <j>.
The differential calculus of forms is based on the exterior derivative d.
For a 0form (function) / on a surface, the exterior derivative is, as before,
the 1form df such that df(\) = vf/]. Wherever forms appear, the exterior
derivative of a pform is a (p + l)form. Thus for surface the only new
definition we need is that of the exterior derivative dtf> of a 1form <f>.
4.4 Definition Let be a 1form on a surface M. Then the exterior de
rivative d<f> of <!> is the 2form such that for any patch x in M ,
d<j>{x u , x v ) = — (<f>(x v )) — — (0(x„)).
du dv
As it stands, this is not yet a valid definition; there is a problem of con
sistency. What we have actually defined is a form d x <f> on the image of each
patch x in M. So what we must prove is that on the region where two patches
overlap, the forms d x <f> and dy<f> are equal. Only then will we have obtained
from <f> a single form dfy on M.
4.5 Lemma Let 4> be a 1form on M. If x and y are patches in M, then
d x <t> = d y <t> on the overlap of x{D) and y(E).
Proof. Because y u and y v are linearly independent at each point, it suf
fices by Lemma 4.2 to show that
(dy<l>) (y u ,y v ) = (dx0)(yu,ye).
Now as in Corollary 3.4, we write y = x(u,v) and, as in an earlier exercise,
deduce by the chain rule that
_ du dv
du du
(1)
_ du dv
dv dv
where x„ and x„ are henceforth evaluated on (u,v). Then by Lemma 4.2,
(40) (y«,y«) = J (d x <i>) (x u ,x v ) (2)
where J is the Jacobian (du/du)(dv/dv) — (du/dv)(dv/du). Thus it is
clear from Definition 4.4 that to prove (dy<j>) (y„,y„) = id^) (y«,y„), all we
need is the equation
± (*.)£<*.) /{£<*.)£ <*.)}. (3)
It suffices to operate on (d/du)(<f>y v ), for merely reversing u and v will
then yield (d/dv)(<f>y u ). Since (3) requires us to subtract these two deriva
Sec. 4] DIFFERENTIAL FORMS ON A SURFACE 155
tives, we can discard any terms that will cancel when u and v are everywhere
reversed.
Applying <f> to the second equation in (1) yields
*(y.) = 0(x„) ^ + tf>(x e ) /.
Hence by the chain rule,
where in accordance with the remark above we have discarded two sym
metric terms. Next we use the chain rule— and the same remark — to get
s^>(s<*)£+)s+fe ( ^s + )s<«
Now reverse u and v in (5) (and also u and v) and subtract from (5)
itself. The result is precisely equation (3). 
It is difficult to exaggerate the importance of the exterior derivative. We
have already seen in Chapter I that it generalizes the familiar notion of dif
ferential of a function, and that it contains the three fundamental deriva
tive operations of classical vector analysis (Exercise 8 of Chapter I, Section
6). In Chapter II it is essential to the Cartan structural equations (Theorem
8.3). Perhaps the clearest statement of its meaning will come in Stokes
theorem (6.5), which could actually be used to define the exterior deriva
tive of a 1form.
On a surface, the exterior derivative of a wedge product displays the
same linear and Leibnizian properties (Theorem 6.4 of Chapter I) as in
E ; see Exercise 3. For practical computations these properties are apt to
be more efficient than a direct appeal to the definition — compare the dis
cussion of the Euclidean case on page 25. Examples of this technique appear
in subsequent exercises.
The most striking property of this notion of derivative is that there are
no second exterior derivatives: Wherever forms appear, the exterior deriva
tive applied twice always gives zero. For a surface, we need only prove this
for Oforms, since even for a 1form <j>, the second derivative d(d<}>) is a
3 form, hence is automatically zero.
4.6 Theorem If / is a (differentiate) realvalued function on M, then
d(df) = 0.
Proof. Let yp = df, so we must show drp = 0. It suffices by Lemma 4.2
to prove that for any patch x in M we have (df) (x u ,x„) = 0. Now using
Exercise 4 of Section 3 we get
156 CALCULUS ON A SURFACE [Chap. IV
*(0 = df(x u ) = x u [f) = % (fx)
du
and similarly
dv
Hence
#(»., x.) = L (**.)  > (+*„) = £<£>  ^M = o 
Many computations and proofs reduce to the problem of showing that
two forms are equal. As we have seen, to do so it is not necessary to check
that the forms have the same value on all tangent vectors. In particular,
if x is a coordinate patch, then
(1) for 1forms on x(D): </> = yp if and only if <£(x„) = \j/{x u ) and
<f>(x v ) = i//(x v );
(2) for 2forms on x(D): \i = v if and only if n(x u , x v ) = v(x u , x v ).
(To prove these criteria we express arbitrary tangent vectors as linear
combinations of x„ and x„.) More generally, x u and x„ maybe replaced by
any two vector fields that are linearly independent at each point.
Let us now check that the rigorous results proved in this section are con
sistent with the rules of operation stated in Chapter I, Section 6.
4.7 Example Differential forms on the plane E 2 . Let U\ = u and u 2 = v
be the natural coordinate functions, and U\,U 2 the natural frame field on
E . The differential calculus of forms on E 2 is expressed in terms of Ui and
Ui as follows :
If / is a function, <j> a 1form, and t\ a 2form, then
(1) <t> — /i dui + f 2 du 2 , where /»• = 4>(Ut).
(2) rj = gduidih, where g = r](Ui,U 2 ).
(3 ) for \[/ = (ft dui + g 2 du2 and <f> as above,
A ^ = C/i02 — fnQi)dui du 2 .
(4) df = M. du, + P du 2 .
dui du 2
(5) d<f> = ( — — J dui du 2 (<f> as above).
\dui du 2 /
For the proof of these formulas, see Exercise 4.
Similar definitions and coordinate expressions may be established on any
Euclidean space. In the case of the real line E 1 , the natural frame field
Sec. 4] DIFFERENTIAL FORMS ON A SURFACE 157
(Definition 2.4 of Chapter I) reduces to the single unit vector field Ui for
which Ui[f] = df/dt. All pforms with p > 1 are zero, and if <j> is a 1form,
then<£ = <f>{Ui) dt.
Some examples of forms will appear in subsequent exercises; however,
many more will occur naturally throughout Chapters VI and VII, where
their properties have direct geometric meaning.
EXERCISES
1. If <f> and ^ are 1 forms on a surface, prove that <j> a rp = — \f/ a <j>.
Deduce that <f> a <j> = 0.
2. A form 4> such that dxf> = is said to be closed. A form <j> such that
<f> = d£ for some form £ is exact. (So if <p is a pform, £ is necessarily a
(p — l)form.) Prove
(a) Every exact form is closed.
(b) No 0form is exact, and on a surface every 2form is closed.
(c) Constant functions are closed 0forms.
3. Prove the Leibnizian formulas
d(fg) = dfg+fdg d(j<f>) = df A <t> + f d*
where/ and g are functions on M, and <f> is a 1form (Hint: By definition
(fo)(v P ) = f(p)<t>(y P ); hence f<f> evaluated on x u is /(x)<£(x„).)
4. (a) Prove formulas (1) and (2) in Example 4.7 using the remark pre
ceding Example 4.7. (Hint: Show (duidu2)(Ui, U 2 ) = 1.)
(b) Derive the remaining formulas using the properties of d and the
wedge product.
5. If / is a realvalued function on a surface, and g is a function on the
real line, show that
y P \g(f)] = g'(f)y P U)
Deduce that
dig if)) =g'(f)dj.
6. If /, g, and h are functions on a surface M, and <t> is a 1form, prove:
(a) d(fgh) = ghdf + fh dg + fg dh,
(b) d(4>f) =fd<i><t> a df (tf = /0),
(c) (df a dg)(v,w) = v[/]w[<7]  v[g]w\f].
7. Suppose that M is covered by open sets lli, • • • , m^, and on each "U;
there is defined a function f { such that/* — /_, is constant on the overlap
158 CALCULUS ON A SURFACE [Chap. IV
of lit and C U 7 . Show that there is a 1form <f> on M such that </> = dfc
on each ll*. Generalize to the case of 1 forms fa such that fa — <f>j is
closed.
8. Let y: E — > M be an arbitrary mapping of an open set of E 2 into a
surface M. If is a 1form on M, show that the formula
r
d<f>(y u , y v ) = — (0 y.)  ^ (4>y«)
is still valid even when y is not regular or onetoone.
(Hint: In the proof of Lemma 4.5, check that equation (3) is still
valid in this case.)
A patch x in M establishes a onetoone correspondence between an
open set D of E 2 and an open set x(D) of M. While we have empha
sized the function x: D —> x(D), there are some advantages to empha
sizing instead the inverse function x _1 : x(D) — > D.
9. If x: D — ► M is a patch in M, let u and be the coordinate functions
of x _1 , so x _1 (p) = (#(p), P(p)) for allp in x(Z>). Show that
(a) u and are differentiable functions on x(D) such that
u(x(u,v)) = w, v(x(u,v)) = v.
These functions constitute the coordinate system associated with x.
(b) du(x u ) = 1 du{x v ) =
dv(x u ) = dv(x v ) = 1.
(c) If ^ is a 1form and 77 is a 2form, then
<t> = f du + g dv where /(x) = <f>(x u ), g(x) = <f>(x v )
r\ = h dudv where h(x) = 77 (x M , x„).
(Hint: For (b) use Ex. 4(b) of IV.3.)
10. Identify (or describe) the associated coordinate system u, v of
(a) The polar coordinate patch x(u, v) = (u cos v, u sin v) defined
on D: u > 0, < v < 2tt.
(b) The identity patch x(u, v) = (u,v) in E 2 .
(c) The geographical patch x in the sphere.
5 Mappings of Surfaces
To define differentiability of a function from a surface to a surface, we follow
the same general scheme used in Section 3, and require that all its coordi
nate expressions be differentiable.
Sec. 5]
MAPPINGS OF SURFACES
159
5.1 Definition A function F: M — > N from one surface to another is dif
ferentiable provided that for each patch x in M and y in N the composite
function y _1 Fx is Euclidean differentiate (and defined on an open set of
E 2 ). F is then called a mapping of surfaces.
Evidently the function y~Vx is defined at all points (u, v) of D such that
F(x(u, v)) lies in the image of y ( Fig. 4.29). As in Section 3 we deduce
from Corollary 3.3 that, in applying this definition, it suffices to check
enough patches to cover both M and N.
5.2 Example (1) Let 2 be the unit sphere in E (center at 0) but
with north and south poles removed, and let C be the cylinder based on the
unit circle in the xy plane. So C is in contact with the sphere along the
equator. We define a mapping F: 2 — * C as follows: If p is a point of 2,
draw the line orthogonally out from the z azis through p, and let F(p) be
the point at which this line first meets C, as in Fig. 4.30. To prove that F
is a mapping, we use the geographical patch x in 2 (Example 2.2), and for
C we use the patch y(w, v) = (cos u, sin u, v). Now
x(u, v) = (cos v cos u, cos v sin u, sin v),
so from the definition of F, we get
F(x(u, v)) = (cos u, sin u, sin v).
But this point of C is y(u, sin v); hence
F(x(u, v)) = y(w, sin v).
Applying y~ to both sides of this equation, we
find
(y  Fx) (u, v) = (u, sin v)
so y~ l Fx is certainly differentiable. (Actually x
does not entirely cover 2, but the missing semicircle
can be covered by a patch like x.) We conclude
that F is a mapping.
FIG. 4.30
160
CALCULUS ON A SURFACE
[Chap. IV
(2) Stereographic projection of the punctured sphere 2 onto the plane.
Let 2 be a unit sphere resting on the xy plane at the origin, so the center of
2 is at (0, 0, 1 ) . Delete the north pole n = (0, 0, 2 ) from 2. Now imagine that
there is a light source at the north pole, and for each point p of 2, let
P(p) be the shadow of p in the xy plane (Fig. 4.31). As usual, we identify
the xy plane with E 2 by (pi, p 2 , 0) <> (pi, p%). Thus we have defined a
function P from 2 onto E 2 . Evidently P has the form
\ r r /
where r and R are the distances fromp and P (p), respectively, to the z axis.
But from the similar triangles in Fig. 4.32, we see that R/2 = r/(2 — p 3 );
hence
P( PllP! , P 3)=(^,^)
\2 — p 3 2 — pzj
Now if x is any patch in 2, the composite function Px is Euclidean dif
ferentiate, so P: 2 — ► E 2 is a mapping.
Just as for mappings of Euclidean space, each mapping of surfaces has a
derivative map.
5.3 Definition Let F: M — > N be a mapping of surfaces. The derivative
map F* of F assigns to each tangent vector v to M the tangent vector F* (v)
to N such that: If v is the initial velocity of a curve a in M, then F* (v) is
the initial velocity of the image curve F(a) in N (Fig. 4.33).
Furthermore, at each point p, the derivative map F* is a linear trans
formation from the tangent plane T P (M) to the tangent plane T F ( P )(N)
(see Exercise 13 ) . It follows immediately from the definition that F* pre
serves velocities of curves: If a = F (a) is the image in N of a curve a in
M, then F* (a) = a'. As in the Euclidean case, we deduce the convenient
property that the derivative map of a composition is the composition of
the derivative maps (Exercise 14).
FIG. 4.31
P(p)
Sec. 5]
MAPPINGS OF SURFACES
161
F*(v)
FIG. 4.33
FIG. 4.34
The derivative map of a mapping F: M — » N may be computed in terms
of partial velocities as follows. If x: D — * M is a parametrization in M, lefr
y be the composite mapping F(x): D —* N (which need not be a parametri
zation). Obviously F carries the parameter curves of x to the corresponding
parameter curves of y. Since F* preserves velocities of curves, it follows at
once that
F*(x u ) = y u F*(x v ) = y„
Since x„ and x„ give a basis for the tangent space of M at each point of
x(D), these readily computable formulas completely determine F*.
The discussion of regular mappings in Section 7 of Chapter I translates
easily to the case of a mapping of surfaces F : M — > N. F is regular provided
all of its derivative maps F* p : T P (M) — * T F ( P )(N) are onetoone. Since
these tangent planes have the same dimension, we know from linear algebra
that the onetoone requirement is equivalent to F* being a linear iso
morphism. A mapping F: M —> N that has an inverse mapping F 1 :^ — > M
is called a diffeomorphism. We may think of a diffeomorphism F as smoothly
distorting M to produce N. By applying the Euclidean formulation of the
inverse function theorem to a coordinate expression y _1 ^x for F, we can
deduce this extension of the inverse function theorem (7.10 of Chapter I).
5.4 Theorem Let F: M — > iV be a mapping of surfaces, and suppose that
F* p : T P (M) — > T F ( P ) (N) is a linear isomorphism at some one point p of M .
Then there exists a neighborhood 11 of p in M such that the restriction of F
to 11 is a diffeomorphism onto a neighborhood V of F(p) in N (Fig. 4.34).
T62
CALCULUS ON A SURFACE
[Chap. IV
/a vgarameter curve 01 y^
a Mparameter curve of^T
FIG. 4.35
An immediate consequence is that a onetoone regular mapping F of M
onto N is a diffeomorphism. For since F is onetoone and onto, it has a
unique inverse function P 1 , and F~ l is a (differentiable) mapping, since
on each neighborhood V, as above, it coincides with the inverse of the
diffeomorphism 11 — » V.
5.5 Example Stereographic projection P: S — > E 2 is a diffeomorphism.
It is clear from Example 5.2 that P is a onetoone mapping from the
punctured sphere S to the plane E 2 . Thus we need only show that P* is
onetoone at each point. A minor modification of the geographical patch
in Example 2.2 yields a parametrization
\(u, v) = (cos v cos u, cos v sin u, 1 + sin v)
of all of 2 except the south pole, located at the origin 0.
Now its geometric definition shows that P carries the wparameter curves
of x (circles of latitude) to circles in the plane, centered at the origin, and
that P carries the vparameter curves (meridians of longitude) to straight
lines radiating out from the origin, as shown in Fig. 4.35.
Indeed these two families of image curves are just the parameter curves
of y = P(x), and from the formula for P in Example 5.2 we get
y(u,v) = P(x(u,v)) =
ft
cos v cos u 2 cos
sin v
1 
v sin u\
sin v )
Since P* (x„) = y„, P* (x„) = y„, the regularity of P* may be proved by
computing y u and y„, which turn out to be orthogonal and nonzero, hence
linearly independent. (At the south pole a different proof is required,
since x is not a parametrization there (see Exercise 15) . ) We conclude that
P is a diffeomorphism.
Differential forms have the remarkable property that they can be moved
from one surface to another by means of an arbitrary mapping, f Let us
experiment with a 0form, that is, a realvalued function /. If F : M —> N is
t This is not the case with vector fields.
Sec. 5]
MAPPINGS OF SURFACES
163
a mapping of surfaces and/ is a function on M, there is simply no reasonable
general way to move / over to a function on N. But if instead/ is a function
on N, the problem is easy; we pull/ back to the composite function f(F) on
M. The corresponding pullback for 1 forms and 2forms is accomplished as
follows.
5.6 Definition Let F: M — > N be a mapping of surfaces.
(1) If <^> is a 1form on N, let F*<f> be the 1form on M such that
(F**)(v) = <t>(F*v)
for all tangent vectors v to M.
(2) If t] is a 2form on N, let F*r] be the 2form on M such that
(F*,)(v,w) = iK*»v,*W)
for all pairs of tangent vectors v, w on M (Fig. 4.36).
When we are dealing with a function / in its role as a 0form, we shall
sometimes write F*f instead of f(F), in accordance with the notation for
the pullback of 1 forms and 2forms.
The essential operations on forms are sum, wedge product, and exterior
derivative; all are preserved by mappings.
5.7 Theorem Let F: M —> N be a, mapping of surfaces, and let £ and r\
be forms on N. Then
(1) F*(Z + r,) = F*Z + F* V ,
(2) F*($ a „) = F*£ a F*n,
(3) F*(dt) = d(F*£).
Proof. In (1), £ and rj are both assumed to be pforms (degree p — 0, 1, 2)
and the proof is a routine computation. In (2), we must allow £ and r\ to
have different degrees. When, say, £ is a function /, the given formula means
simply F*(fr)) = f(F)F*(ri). In any case, the proof of (2) is also a straight
forward computation. But (3) is more interesting. The easier case when £
is a function is left as an exercise (Exercise 8 ) , and we address ourselves to
the difficult case when £ is a 1form.
F*(w)
FIG. 4.36
164 CALCULUS ON A SURFACE [Chap. IV
It suffices to show that for every patch x: D — > M
(d(F*^))(x u ,x v ) = {F*(di))(x u , x v ).
Let y = F(x), and recall that F*{x u ) = y M and F*(x v ) = y v . Thus using
the definitions of d and F*, we get
d(F*i;)(x u ,x,) = ±{(F*t)(x,)}  1 {(F*Z)(x u )}
du OV
 k ((( *' )}  1 (i(y  )) 
Even if y is not a patch, Exercise 8 of Section 4 shows that this last expres
sion is still equal to di(y u , y v ). But
dtiju, y„) = dit(F*x u , F*x v ) = (F*(d£))(x u , x v ).
Thus we conclude that d (F*£) and F* (d£) have the same value on x u , x v . 
The elegant formulas in Theorem 5.7 are the key to the deeper study of
mappings. In Chapter VI we shall apply them to the connection forms of
frame fields to get fundamental information about the geometry of mappings
of surfaces.
EXERCISES
1. Let M and N be surfaces in E 3 . If F: E 3 — > E is a mapping such that
the image F(M) of M is contained in N, then the restriction of F to
M is a function F \ M : M — > N. Prove that F \ M is a mapping of sur
faces. {Hint: Use Theorem 3.2.)
2. Let S be the sphere of radius r with center at the origin of E . Describe
the effect of the following mappings F: S — *■ 2 on the meridians and
parallels of S.
(a) F(p) = p. (b) F(pi,p2,pz) = (ps, Pi, Ps).
f \ Vf \ (Vl + V* Pi ~ P2 \
(c) F{p h p 2 , p,) = I ^ ' v/2 ' ~ P 7
Let M be a simple surface, that is, one which is the image of a single
proper patch x : D — > E 3 . If y : D — > ]V is any mapping into a surface N,
show that the function F: M —> N such that
F(x(u, v)) = y(w, y) for all (u, v) in D
is a mapping of surfaces. (Hint: Write F = yx _1 , and use Corollary 3.3.)
Sec. 5] MAPPINGS OF SURFACES 165
4. Use Exercise 3 to construct a mapping of the helicoid H (Ex. 7, IV.2)
onto the torus T (Example 2.6) such that the rulings of H are carried
into meridians of T.
5. If 2 is the sphere  p  = r, the function A : 2 — > 2 such that J. (p) = — p
is called the antipodal mapping of 2. Prove that A is a diffeomorphism
and that A* (v p ) = ( — v)_ p .
6. Let x: D — > M be a coordinate patch in a surface M. For any form
on M, the form x*(0) on D is called a coordinate expression for 0.
(When^ is aOform, that is, a function, then x*(0) = ^(x), so this
terminology is consistent with that of IV.3.)
If <j> is a 1form and v a 2form, prove
(a) x*(<£) = <f>(x u )du + <j>(x v )dv. (b) x*(v) = p(x M , x v )dudv.
(c) x*(<ty) = (A (0 x„)  A (0 xj") dw dw.
(In practice, instead of substituting in the formula (c), it is usually
easier to apply the exterior derivative to the formula (a).)
7. (Continuation). Let x be the geographical patch in the sphere 2.
(a) If <f> is the 1form on 2 such that <f>(\ p ) = piv 2 — p 2 v u show that
0(x u ) = r 2 cos 2 y and <f>(x v ) = 0. Then find the coordinate expres
sions for <(> and a\f>.
(b) Prove that the formula v(\ p , w p ) = p$(viw 2 — v 2 Wx) defines a 2
form on 2 and find its coordinate expression.
8. Let F: M —> N be a mapping, and g a function on N.
(a) Prove that F preserves directional derivatives in this sense: If v
is a tangent vector to M, then y\g(F)] = (F*\)\g].
(b) Deduce that F*(dg) = d(F*(g)).
9. If x: D — ► M is a parametrization, prove that the restriction of x to a
sufficiently small neighborhood of a point (wo, v ) in D is a patch in M.
(Thus a parametrization may be cut into patches.)
10. If G: P — > M is a regular mapping onto Af, and #: P — ► iV is an arbitrary
mapping, then the formula F(G(p)) = #(p) is consistent provided
G(p) = G(q) implies H(p) = #(q) for all p, q in P. Prove that in
this case F is a welldefined (differentiable) mapping.
M >N
G\ /H
P
We shall frequently apply this result in the case where G is a para
metrization of M.
166 CALCULUS ON A SURFACE [Chap. IV
11. Let x: E 2 —*■ T be the parametrization of the torus given in Example 2.6.
In each case below, show that the formula
F(x(u, v)) = x(f(u, v), g(u, v))
is consistent (Exercise 10), and describe the resulting mapping
F:T^>T. (For example, give its effect on the meridians and parallels
of T.)
(a) / = 3m, g = v. (c) / = v, g = u.
(b) / = u + x, g = v + 2tt. (d) / = u .+ v, g = u  v.
Which of these mappings are diffeomorphisms?
12. Let F: M* N be a mapping. Let x be a patch in M and let y = F(x).
(Note that y lies in N, but need not be a patch. ) If
a(t) = x(a t (t), oa(0)
is a curve in M, prove that the image a = F(a) in N has velocity
_/ dai i s dat , ■,
oc = jT y«(ai, as) + j7 y.voi, a 2 ).
13. Deduce from Exercise 12:
(a) The invariance property needed to justify definition (5.3) of F*.
(b) The fact that derivative maps F*:T P (M) — > T F ( p) (N) are linear
transformations .
" w c
14. Given mappings M > N > P, let GF: M ^ P be the composite
function. Show that
(a) GF is a mapping, (b) (GF)* = G*F*, (c) (GF)* = F*G*,
that is, for any form £ on N, F*(G*£) = (GF)*(£). (Note the reversal
of factors for (GF)*; forms travel in the opposite direction from points
and tangent vectors. )
15. For stereographic projection P: 2 — » E 2 , show that the derivative map
at the origin is essentially just an identity mapping. (Hint: Express
P near in terms of a Monge patch. )
16. (a) Prove that the inverse mapping of stereographic projection
P: 2 — » E 2 is given by the formula
p (m , ,) = (^iil ,
where / = w 2 + v. (Show that both PP~ l and P~ l P are identity
mappings. )
(b) Deduce that the entire sphere 2 can be covered by only two
patches. (The scheme in IV.l requires six.)
See. 6] INTEGRATION OF FORMS 167
6 Integration of Forms
Differential forms have yet another role in the calculus, which the reader
probably noticed when they first appeared in Chapter I. In, say, a double
integral fff(u, v) du dv, it is a 2form on E 2 that appears after the integral
signs. In a sense, it is only on Euclidean space that forms are actually
integrated. But we can easily extend this notion of integration to forms on
an arbitrary surface — by pulling them back to Euclidean space and then
integrating.
Consider first the onedimensional case. By a curve segment (or lseg
ment) in a surface M we shall mean a "curve" a: [a, b] — > M defined on a
closed interval in the real line E 1 . (Differentiability for a means that it
can be extended to a genuine curve on a larger open interval as required by
Definition 4.1 of Chapter I.)
Now let <f> be a 1form on M. The pullback a*4> of </> to the interval
[a, b] has the expression /(f) dt, where by the remarks following Example
4.7,
f(t) = (a**)(tfi(0) =*(*»(tfi(0)) =<*>(«' (0).
Thus the scheme described above yields the following definition.
6.1 Definition Let <£ be a 1form on M, and let a: [a, 6]^Mbea 1 seg
ment (Fig. 4.37). Then the integral of $ over a is
f $ = / a*4> = f <f>(a(t))
dt
In engineering and physics, the integral f a <f> is called a line integral, and
it has a wide variety of uses. For example, let us suppose that a vector field
V on a surface is a force field, so for each point p of M, V(p) is a force
exerted at p. Returning to our original idea of curve, we further suppose
that a: [a, b] — » M describes the motion of a mass point — with a(t) its
position at time t. What is the total amount of work W needed to move a
from p = a (a) toq = a (b)? The discussion of velocity in Chapter I,
graph <£(«')
FIG. 4.37
168
CALCULUS ON A SURFACE
[Chap. IV
Section 4, shows that for At small, the route of the curve a from a(t) to
a(t + At) is approximately the straight line segment described by At a(t).
Now the moving point is opposed only by the component of force tangent
to a, that is,
V(a)
= \\V(a)  cos *
(Fig. 4.38). Thus the work done against the force during time At is (ap
proximately) force V (a (0). [a (t)/\\ d (t)\\ ] times distance a'(0A*.
Adding these contributions over the whole time interval [a, b] and taking
the usual limit, we get
W
= f V(a(t))a'(t)
"a
dt.
To express this more simply, we introduce
the dual 1form <j> such that for each tangent
vector w at p, <*>(w) = wF(p). Then, by
Definition 6.1, the total work is just
a(t + At)
w
■/.
We emphasize that this notion of line in
tegral — like everything we are doing with
fig. 4.38 forms — applies without change if the surface
M is replaced by a Euclidean space or, in
deed, by any manifold (Section 8).
When the 1form <j> is an exterior derivative df, the line integral J a </> has
an interesting property which generalizes the fundamental theorem of
calculus.
6.2 Theorem Let/ be a function on M , and let a: [a, b] — » M be a 1seg
ment in M fromp = a(a) toq = a(b). Then
Proof. By definition,
f df = /(q)  /(p).
•'a
/ df = f df(a') dt.
"a "a
But
df(a') = a'lfl =  (fa).
Sec. 6} INTEGRATION OF FORMS 169
Hence by the fundamental theorem of calculus,
L df = f.% (fa) dt = /(a(6)) _ /(a(a)) = /(q) " /(p)  ■
The integral j a df is thus said to be pathindependent. In the language
used above, if the force field V has dual 1form df, the work done depends
not on where the point a (t) moves, but only on where it starts and finishes.
In particular, if it follows a closed curve, p = a (a) = a(b) = q, it does no
(total) work at all.
Mathematically we look at the preceding theorem roughly as follows:
the "boundary" of the segment a from p to q is q — p, where the purely
formal minus sign indicates that a goes from p and to q. Then the integral
of df over a equals the "integral" of / over the boundary q — p; that is,
/(q) — /(p). This interpretation will be justified by the analogous theorem
(6.5), in dimension 2.
Now a twodimensional interval is just a closed rectangle R : a ^ u ^ b,
c ^ v ^ d in E 2 . And a 2segment in M is a differentiate mapping x: R —> M
of a closed rectangle into M (Fig. 4.39 ) . (As before, differentiability means
one can extend x differentiably to an open set containing R. )
Although we use the patch notation x, we do not assume that x is either
regular or onetoone. The partial velocities x u and x„ are still available,
however, even when x is not a patch.
If 77 is a 2form on M, then the pullback x*tz of 77 has, using Example 4.7,
the coordinate expression h du dv, where
h = (x*?/)(£/i, C/ 2 ) = rj(x*C/i, x*C/ 2 ) = r?(x M , x„).
Thus by strict analogy with Definition 6.1 we establish
6.3 Definition Let tj be a 2form on M , and let x: R —> M be a 2segment.
Then the integral of rj over x is
JJ v = JJ x * v = J J v ^ Xu ' Xv ^ du dv '
The physical applications of this notion of integral are perhaps richer
FIG. 4.39
170
CALCULUS ON A SURFACE
[Chap. IV
than those of Definition 6.1, but we must proceed without delay to the
twodimensional analogue of Theorem 6.2.
6.4 Definition Let x: R — > M be a 2segment in M with R the closed rec
tangle a ^ u ^ b, c ^ v ^ d (Fig. 4.40). The edge curves (or edges) of x
are the 1 segments a, (3, y, 8 such that
a(u) = x(u, c)
0(v) = x(b,v)
y(u) = x(u, d)
8(v) = x(a, v)
d
(y)
■ (*) R (/?)
IS
a b "
FIG. 4.40
Then the boundary dx of the 2segment x is the formal expression
dx = a + P — y — 8.
These four curve segments are what we get by considering the function
x:R+M only on the four line segments that comprise the boundary of the
rectangle R. The formal minus signs before y and 8 in dx remind us that y
and 8 must be "reversed" to give a consistent trip around the rim of R, and
thus of x (Fig. 4.41). Then if <f> is a 1form on M, the integral of </> over the
boundary of x is defined to be
/0=/*^+/^/'0/ , 0.
J dx Ja J B J", >&
J dx J a •'/S
The twodimensional analogue of Theorem 6.2 is then
6.5 Theorem (Stokes' Theorem) If <f> is a 1form on M, and x: R — > M
is a 2segment, then
II d<f> = /
Proof. We shall work on the double integral and show that it turns out
R
FIG. 4.41
Sec. 6]
INTEGRATION OF FORMS
171
to be the integral of <f> over the boundary of x. Combining Definitions 6.3
and 4.4 we have
J J d<f> = J J (d4>) (x« , x„) du dv = J J (— (<f>x v )  — (<t>x u ) J du dv
Let/ = <t>(x u ) and g = <f>(x v ); this equation then becomes
[f d4= (f d ± du dv  [f d 4du dv (1)
JJ X JJr du JJr dv
Now we treat these double integrals as iterated integrals. Suppose the
rectangle R is given by the inequalities a^u^b, c^v^d. Then inte
grating first with respect to u, we find
[[ ^ dudv = [ I(v)dv, where I(v) = [ ^ (u, v) du.
JJr dU J c J a dU
In the partial integral defining I (v), v is constant, so the integrand is just
the ordinary derivative with^ respect to u. Thus the fundamental theorem
of calculus applies to give
I(v) = flf(M)  9(a,v).
(Fig. 4.42). Hence
[[ d 4 dudv = [ g(b, v) dv  f g(a, v) dv. (2)
JJr dU J c J c
Again we work on the first integral. Now by definition
g(b,v) = 0(x„(6, v)).
But x v (b, v) is precisely the velocity /3'(y) of the "right side" curve in
dx. Hence by Definition 6.1,
J g(b,v)dv = f 4>(p'(v))(kr == / <f>.
(a,v)
(b, v)
FIG. 4.42
172 CALCULUS ON A SURFACE [Chap. IV
A similar argument shows that the second integral in (2) is f& <f>; hence
ff m £* , *f,*f,+ (3)
In the same way — but integrating first with respect to v\ — we find
//.s**/ T */.* (4)
Assembling the information in (1), (3), and (4) we obtain the required
result:
//.*{i* /*}"{/,♦ /.♦}"/.♦
i
Stokes' theorem may be considered a twodimensional formulation of
the fundamental theorem of calculus; it ranks as one of the most useful
results in mathematics. Alternative formulations of the theorem and ex
tensive applications may be found in texts on advanced calculus or applied
mathematics; we shall use it to study the geometry of surfaces.
The line integral /« is not particularly sensitive to reparametrization of
a; all that matters is the direction in which the route of a is traversed.
The following lemma uses the notation of Exercise 10, II.2.
6.6 Lemma Let a (h) : [a, b] — > M be a reparametrization of a curve seg
ment a: [c, d] — ► M . For any 1form <f> on M
/ ♦
Ja(h)
I <j) if h is orientationpreserving
•'a
I 4> if h is orientationreversing
J a
Proof. Since a(h) has velocity
a(h)' = ja'(h),
we have
dh
f <t> = f 4>(a{hY) du= f 0(«'(A))
1 du
du
Now we use the theorem on change of variables in an integral. If h is
orientation preserving, then h(a) = c and h(b) = d, so the integral above
becomes
/ 4>(a) du = I <j>.
Sec. 6] INTEGRATION OF FORMS 173
But in the orientationreversing case, h(a) = d and h(b) = c, which gives
] <*>(«') du = ] <f>(a) du = J 4> I
This lemma lets us give a concrete interpretation to the formal minus
signs in the boundary dx = a + & — 7 — 5 of a 2segment x. For any
curve £: [t , U] — > M, let — £ be any orientationrevemngf reparametrization
of £, say
Thus by the lemma,
(0(0 = €(*> + <i  0.
and if x is a 2segment, then
[<l>=l<t>+[<1>f<l>l<t>=l<f>+f<t>+[ <{>+ [ <f>.
J dx J a J0 Jy Ji J a J P Jy JS
EXERCISES
1. If a = (ai, a 2 ) is a curve in E 2 and is a 1form, prove this computa
tional rule for finding <f>(a) dt: Substitute u = ai(t) and v = a 2 (t)
in a coordinate expression <f> = / (u, v) du + g (u, v) dv.
2. Consider the curve segment a: [— 1, 1] — > E 2 such that a(t) = (t,t 2 ).
(a) If <f> = v du + 2wi; dv, compute /« 0.
(b) Find a function / such that df = <j> and check Theorem 6.2 in this
case.
3. Let be a 1form on a surface M.
(a) If <f> is closed, show that j dx </> = for every 2segment in M .
(b) If is ezad, show (more generally) that
s/.*°
for any "cycle" of curve segments «i, • • • ,
a k (<xk+i = cti ) such that a; ends at the start
ing point of cti+i (Fig. 4.43). (Closed means
d4> = 0, ezaef means = d/ ; see Ex. 2 of
IV.4.)
4. The 1form
FIG. 4.43
174 CALCULUS ON A SURFACE [Chap. IV
_ u dv — v du
* ~ U 2 + V 2
is welldefined on the plane E 2 with the origin (0, 0) removed. Show:
(a) yp is closed, but not exact. {Hint: Integrate around the unit circle
and use Exercise 3.)
(b) if \f/ is restricted to, say, the right half plane, u > 0, then \]/ be
comes exact.
5. (Continuation) . It follows from Ex. 12 of II. 1 that every curve a in
E that does not pass through the origin can be written in the polar
form
a(t) = (r(t) cos #(t),r(t) sintK*)).
Prove that for every closed curve a, (l/2ir) J a ^ is an integer. This
integer is called the winding number of a; it represents the total alge
braic number of times a has gone around the origin in the counter
clockwise direction.
6. Let x be a patch in a surface M. For a curve segment
a(t) = x(ai(0, O2(0), a ^ t ^ b,
show that
/.♦/>*•>£+♦<*>£)*
where x u and x v are evaluated on (a iy a 2 ). (This generalizes Ex. 1, since
we can use the identity patch x(w, v) = (u, v) on E .)
7. Let a be the closed curve
a(t) = x(mt, nt), S t S 2x
in the torus T (see Ex. 11 of IV.3.) Compute
(a) Ja £, where £ is the 1form on T such that £(x„) = 1, and £(x c ) = 0.
(b) /« r?, where ?7 is the 1form such that rj (x u ) = and t}(x v ) = 1.
If 7 is an arbitrary closed curve, / 7 £ /2ir gives the total number of
times 7 travels around the torus in the general direction of the parallels,
while j y r)/2ir gives a similar measurement for the direction of the
meridians. This suggests the commonly used notation £ = d&, t\ = d<p,
where t? and <p are the (multivalued ! ) longitude and latitude functions
on T; however, see Exercise 13.
8. Let F: M — ► N be a mapping. Prove:
(a) If a is a curve segment in M and ^ is a 1form on N, then
J a J F(a)
Sec. 6] INTEGRATION OF FORMS 175
(b) If x is a 2segment in M and v is a 2form on N, then
f F*p = [ v.
Jj. J F(x)
9. Let x: R > 2 be the 2segment in the sphere 2 obtained by restricting
the geographical patch
x(w, v) = (r cos v cos u, r cos t; sin u, r sin v)
to the rectangle R: ^ u, v ^ x/2. Find explicit formulas for the edge
curves a, j8, 7, 5 of x, and show these curves and the image x(R) on a
sketch of 2.
10. Let x: R » If be a 2segment denned on the rectangle
#:0 ^ w,v ^ 1.
If <£ is the 1form on M such that
<t>(x u ) = u + v and <£(x„) = wv,
compute jj x d4> and J 3x </> and check the results by Stokes' theorem.
{Hint: x*(cty) = d(x*0) = (»  l)du d».)
11. Same as Exercise 10, except that R: ^ u ^ tt/2, ^ v ^ x, and
0(x u ) = u cos v, <£(x,,) = v sin w.
12. A closed curve a in M is homotopic to a constant provided there is a
2segment x: R > M for which (a) a is, in fact, the a edge curve of x,
(b) = 5, and (c) 7 is a constant curve (Fig. 4.44). (Suppose
R:a ^ u ^ b, c ^ v ^ d. Then as v varies from c to d, the closed
wparameter curve, v = vo, of x varies smoothly from a to the constant
curve 7.) Prove that every closed curve in E 2 is homotopic to constant.
^
FIG. 4.44 FIG. 4.45
13. Let <j> be a closed 1form and a a closed curve. Prove that J a <£ =
if either
(a) <f> is exact, or (b) a is homotopic to a constant.
Deduce that on the torus T meridians and parallels are not homotopic
to constants, and the closed forms £ and 77 (Exercise 7) are not exact.
176 CALCULUS ON A SURFACE [Chap. IV
A surface M for which every closed curve is homotopic to a constant
is said to be simply connected. Thus the plane is simply connected
(Ex. 12 ), but a torus, or a plane with even a single point removed, is not
(Ex. 7 and 5). Roughly speaking, a simply connected surface has no
holes in it, and any four curves a, ft 7, 8 linked together as in Fig. 4.45
are in fact the boundary curves of some 2segment.f Use this assertion
to prove:
14. On a simply connected surface, the integral of a closed 1form <f> is path
independent. (That is, /„ 4> is the same for all a. with the same end
points. )
15. On a simply connected surface, every closed
1form is exact. (Hint: Fix a point p in M and
define /(p) = j s <f> for any curve segment
frompo to p (Fig. 4.46). To show df(y) =
4>(y) for a tangent vector v at p, prove that
if a is a curve with initial velocity v, then
P = a(0)
/(«(*)) =/(p) + f <t>{*\u))du)
FIG. 4.46
7 Topological Properties of Surfaces
We now discuss some of the very basic properties that a surface may possess.
7.1 Definition A surface M is connected provided that for any two points
p and q of M there is a curve segment in M from p to q.
Thus a connected surface M is all in one piece, since one can travel from
any point in M to any other without leaving M . Most of the surfaces we
have discussed so far have been connected; the surface M:z 2 — x 2 — y 2 = 1
(hyperboloid of two sheets) is not connected. Connectedness is a mild and
reasonable condition and might well be included in the definition of surface.
7.2 Definition A surface M is compact provided that M can be covered
by the images of a finite number of 2segments in M .
Roughly speaking, compactness means that the surface is finite in size.
For example, spheres are compact, since if we use the formula for x(u, v)
in Example 2.2 on the closed rectangle
R: ir ^ u ^ 7T, tt/2 S v S x/2,
t For a systematic account of simple connectedness, see pp. 157165 of Lef schetz
[8], where it is shown that spheres are simply connected.
Sec. 7] TOPOLOGICAL PROPERTIES OF SURFACES 177
then 2 is the image of this single 2segment. Similarly the torus of revolu
tion (Example 2.6), or any closed surface of revolution, is compact.
The proof of the following lemma uses this fundamental fact : If / is a
continuous realvalued function defined on a closed rectangle
R: a rg u ^ b, c ^ v ^ d,
then / takes on its maximum at some point of R.
7.3 Lemma If / is a continuous function on a compact surface M, then
/ takes on maximum at some point of M. (Obviously we can also replace
maximum by minimum.)
Proof. By definition there exist a finite number of 2segments
Xi.Ri^M (1 ^ i S k)
whose images cover all of M . Since each x, is differentiate, it is also con
tinuous, so each composite function /x» : R % ■— > R is continuous. Thus by the
remark above, for each index i, there is a point (m», «,) in Ri where the
function /x t takes its maximum. Let, say, /(xi(tti, Vi)) be the largest of
this finite number A: of maximum values. We assert that / takes on its
maximum value at the point m = xi(wi, Vi). In fact, we shall prove that if
p is any point of M, then /(m) ^ /(p). Since the 2segments x x , • • • , x k
cover M , there is an index i such that p = Xj(w, v). But then by the pre
ceding construction,
/(m) = /(xx(M!, v^) ^ f(xi(Ui, Vi)) ^ f(xi(u, v)) = /(p). 
This very useful result can be applied to prove noncompactness. For
example, no cylinder C (as in Example 1.5) is compact, since the coor
dinate function z on C gives the height z(p) of each point p above the xy
plane, and thus has no maximum value on C.
However, Definition 7.2 is a little trickier than it looks. Consider, for
example, the open unit disc D : x 2 + y 2 < 1 in the xy plane. Now, D is a
surface and has finite area x. But D is not compact : It suffices to note that
the continuous function / = (1 — x 2 — y 2 )~ l does not have a maximum on
D. In general, a compact surface cannot have any open edges, as D does.
It must be smoothly closed up everywhere — as well as finite in size — like
a sphere or torus.
Roughly speaking, an orientable surface is one that is not twisted. Of
the many equivalent formulations of orientability the one that follows is
perhaps the simplest.
7.4 Definition A surface M is orientable provided there exists a 2form y.
on M which is nonzero at each point of M.
(A 2form is zero at a point p if it is zero on every pair of tangent vectors
178 CALCULUS ON A SURFACE [Chap. IV
at p. ) Thus the plane E 2 is orientable, since du dv is a nonvanishing 2form.
Although simple, this definition is somewhat mysterious, so we shall prove
a more intuitive criterion,
7.5 Theorem A surface M in E 3 is orientable if and only if there exists a
normal vector field Z on M that is nonzero at each point of M.
Proof. We use the cross product of E 3 to convert normal vector fields into
2forms, and vice versa. For Z as above, define a 2form n on M as follows:
For any pair v, w of tangent vectors to M at p, let
m(v, w) = Z(p)»v X w.
Standard properties of the cross product show that \i is, in fact, a nonvanish
ing 2form on M. Thus M is orientable.
Conversely, suppose that M is orientable, with n a nonvanishing 2form.
If v, w is a linearly independent pair of tangent vectors at p, then
M (v, w) *0,
for otherwise, n would be zero at p. Now define
v X w
Z(p) =
/*(v, w) *
This formula has the remarkable property that it is independent of the
choice of v, w at p. Explicitly, for any other such pair v, \v, it follows from
Lemma 4.2 and the analogous formula for cross products that
v X w _ v X w
/«(v, w) m (v, w) '
We have thus obtained a welldefined Euclidean vector field on all of M .
Again the properties of the cross product show that Z is everywhere normal
to M, but never zero. 
Thus it follows from Lemma 3.8 that every surface in E 3 that can be
defined implicitly is orientable. For example, all cylinders, surfaces of
revolution, and spheres (in fact, all quadric surfaces) are orientable. How
ever, nonorientable surfaces do exist in E 3 . The simplest example is the
famous Mobius band M , which can be made from a strip of paper by giving
it a half twist, then gluing its ends together. (The formal construction of a
particular Mobius band is given in Exercise 7.) M is nonorientable, since
every normal vector field Z on M must somewhere be zero. To see this,
let 7 be a closed curve as indicated in Fig. 4.47, with 7(0) =7(1) = p. If
we assume that Z is never zero, then the twist in M forces the contradiction
Z(7(l)) = —Z(y(0)), since the function t — > Z(y(t)) is differentiate
(that is, Z varies smoothly as it moves around 7).
Sec. 7]
TOPOLOGICAL PROPERTIES OF SURFACES
17?
The three properties discussed in this section — connectedness, compact
ness, and orientability — are topological properties: It is possible to define
them using only open sets and continuous functions, with no differentiabil
ity considerations at all. Using these more general definitions, a more con
ceptual proof can be given of the following result.
7.6 Theorem Let M and N be surfaces in E such that M is contained in
N. If M is compact, and N is connected, then M = N.
(If N were not connected, it might consist of two surfaces, of which M
is one. Similarly the result fails if M is not compact; consider the case of
an open disc M in the xy plane N. )
Proof. Exercise 15 of Section 3 shows that M is an open set of N. We
assume that M does not fill all of N, and deduce a contradiction. By assump
tion there is a point n of N that is not in M. Let m be a point of M . Since
N is connected, there is a curve segment a in N from
a(0) =m
to
:(D =n.
Let t be the least upper bound of those numbers t such that a(t) is in M
We assert that the point p* = a(t*) is in M (Fig. 4.48).
To prove this, consider the realvalued function f on M such that at each
Z(y(0))
Z(y(D)
FIG. 4.48
1 80 CALCULUS ON A SURFACE [Chap. IV
point p of M, /(p) is the Euclidean distance d(p*, p) from p* to p. Now
/ ^ is continuous, in the sense that for each patch x in M the composite
function /(x) is continuous. Since M is compact, Lemma 7.3 applies to
show that / takes a minimum at some point of M. By definition of least
upper bound, there are numbers t < t*, arbitrarily close to t*, such that
a(t) is in M. Since a is continuous (being differentiable) the corresponding
distances d(p*, a(t)) become arbitrarily small; hence the minimum value
of / can only be zero. Thus the only possible point at which / can be a
minimum is p* itself, which means that p* is in the domain M of /.
Since M is an open set of N, it follows that every point of N near enough
to p* is also in M. Thus if t > t* is near enough to t*, a(t) must still be
in M — a contradiction to the definition of t*. 
EXERCISES
1. Decide which of the following surfaces are compact and which are
connected :
(a) A sphere with one point removed. (c) The surface in Fig. 4.10.
(b) The region z > in M: z = xy. (d) M : x 2 + y + z 6 = 1.
(e) A torus with the curve a (t) = x(t, t) removed. (See Ex. 2 of IV.3.)
2. Let Fbea mapping of a surface M onto a surface N. Prove:
(a) If M is connected, then N is connected.
(b) If M is compact, then N is compact.
3. Let F: M — > N be a regular mapping. Prove that if N is orientable>
then M is orientable.
4. Let / be a differentiable realvalued function on a connected surface M.
Prove that:
(a) If df = 0, then / is constant.
(b) If/ is never zero, then either/ > or/ < 0.
5. (a) Prove that a connected orientable surface has exactly two unit
normal vector fields, which are negatives of each other. We denote
these by ±U. (Hint: Use Ex. 4.)
(b) If M is a nonorientable surface, prove that any point of M is
contained in a connected orientable region. (Thus, even on a
nonorientable surface, unit normal vector fields exist locally.)
6. Let F: M — > N be a regular mapping. Prove this generalization of
Theorem 7.6: If M is compact and N is connected, then F carries M
onto N.
7. A Mobius band M (Fig. 4.47) can be constructed as a ruled surface
x(u, v) = j8(m) + v&(u), % ^ v ^ \,
Sec. 8] MANIFOLDS 181
where
/3(w) = (cos u, sin u, 0)
a(«) = (cos)/3( W ) + (sin)t/ 3 .
(The ruling L makes only a half turn as it traverses the circle once. )
(a) Compute
E = V  + [1 + v cos (u/2)] 8 , F = 0, G = 1,
4
and deduce as in Exercise 2 of Section 2 that x is regular.
(b) Show the wparameter curve, v = £, on a sketch of M. Prove that
the wparameter curves are closed and (0 excepted) have period
4tt.
8. Let M * be the surface obtained by removing the central circle j8 from
the Mobius band in the last exercise. Is M* connected? Orientable?
9. (Counterexamples). Give examples to show that the following are all
false :
(a) Converses of (a) and (b) of Exercise 2.
(b) Exercise 3 with F not regular.
(c) Converse of Exercise 3.
10. A surface M in E 3 is closed in E 3 provided the points of E 3 not in M
constitute an open set of E 3 . (If p is not in M, there is an eneighborhood
of p that does not meet M. ) Show that :
(a) Every surface in E 3 that can be described in the implicit form
M : g = c is closed in E .
(b) Every compact surface in E is closed in E .
11. (Boundedness). A surface M in E 3 is bounded provided there is a num
ber R such that  p  ^ R for all points p of M. (Thus M lies inside a
sphere.) Prove that a compact surface in E is bounded.
The last two exercises have shown that a compact surface in E 3 is closed
and bounded; the converse follows from a fundamental topological theorem.
12. Prove Theorem 7.6 assuming M is merely closed in E 3 (instead of
compact).
13. In each case, decide whether the surface M: g = 1 is compact, or
connected :
(a) g = x 2  y s + z\ (c) g = £ + xy\
(b) g = x 4  y + z. (d) g= (x 2 + y 2  4) 2 + (z  4) 2 .
1 82 CALCULUS ON A SURFACE [Chap. IV
14. Prove that every surface of revolution M is connected and orientable,
but that M is compact if and only if its profile curve C is closed.
8 Manifolds
Surfaces in E are a matter of everyday experience, so it is reasonable to
try to investigate them mathematically. But examining this concept with
a critical eye, we may well ask if there could not be surfaces in E 4  • • or
E n • • • or even surfaces that are not in any Euclidean space at all. To devise
a definition for such a surface, we must rely not on our direct experience of
the real world, but on our mathematical experience of surfaces in E 3 . Thus
we shall strip away from the basic definition (1.2) every feature that
involves E in any way. What is left will be just a surface.
To begin with, a surface will be a set M: a collection of any objects
whatsoever, not necessarily points of E 3 . An abstract patch in M will now be
just a onetoone function x: D — > M from an open set D of E 2 into the set
M. There is, as yet, no way to say what it means for such a function to be
differentiable. But all we need to get a workable definition of surface is the
smooth overlap condition (Corollary 3.3). To prove this now is a logical
impossibility, so in the usual fashion of mathematics, we make it an axiom.
8.1 Definition A surface is a set M furnished with a certain collection (P
of abstract patches in M such that:
(1 ) The images of the patches in the collection (P cover M.
(2 ) For any two patches x, y in the collection (P, the composite functions
y _1 x and x _1 y are Euclidean differentiable (and defined on open sets of E 2 ).
This definition generalizes Definition 1.2: A surfaceinE 3 is a surface.
But there are vast numbers of surfaces that can never be found in E 3 .
8.2 Example The projective plane 2. Starting from the unit sphere 2
in E , we construct the projective plane 2 by identifying antipodal points of
2; that is, by considering p and — p to be the same point (Fig. 4.49). For
FIG. 4.49
Sec. 8] MANIFOLDS 183
mally, this means the set 2 consists of all antipodal pairs {p, — p} of points
of the sphere. (Order is not important here; that is, {p, — p} = { — p, p}.)
To make 2 a surface — and later to study it — we use two functions: the
antipodal mapping A(p) = — p of the sphere 2, and the projection P(p) =
{p, p} of 2 onto 2. Note that PA = P, and (J): P(p) = P(q) if and
only if either q = p or q = —p.
Let us call a patch x in 2 "small" if the Euclidean distance between any
two of its points is less than 1. If x: D — ► 2 is a small patch, then the com
posite function P(x) : D — > 2 is onetoone, and is thus an abstract patch
in 2. The collection of all such abstract patches makes 2 a surf ace — the first
condition in Definition 8.1 is clear, and we merely outline the proof of the
second.
Suppose that P(x) and P(y) overlap in 2; that is, their images have a
point in common. If x and y overlap in 2, show that (Py) _1 (Px) = y 1 x,
which by Corollary 3.3 is differentiable and defined on an open set. (Hint:
Use smallness and %.) On the other hand, if x and y do not overlap, replace
y by A(y). Then x and ^4.(y) do overlap, so the previous argument applies.
Conclusion: The projective plane 2 is a surface. f
To emphasize the distinction between a surface in E 3 and the general
concept of surface defined above, we shall sometimes call the latter an
abstract surface. Note that E 2 is an abstract surface if it is furnished with the
single patch x(u, v) = (u, v).
To get as many patches as possible in an abstract surface M, it is custom
ary to enlarge the given patch collection (P to include all abstract patches
in M that overlap smoothly with those in (P. In working with M, these
are the only patches we can use. We emphasize that abstract surfaces Mi
and M 2 with the same set of points are nevertheless different surfaces if
their (enlarged) patch collections (Pi and (P 2 are different.
There is essentially only one problem to solve in establishing the calculus
of an abstract surface M, and that is to define the velocity of a curve in M.
For everything else — differentiable functions, curves themselves, tangent
vectors, tangent vector fields, differential forms, and so on — the definitions
and theorems given for surfaces in E 3 apply without change. (It is neces
sary to tinker with a few proofs to take care of the new Definition 8.3, but
no serious problems arise.) The velocity of a curve fails us in the abstract
case, since before it consisted of tangent vectors to E 3 , and now E 3 is gone.
It makes not the slightest difference what we define the velocity a (t)
to be — provided the new definition leads to the same essential properties
as before. The directional derivative property (Lemma 4.6 of Chapter I)
is what is needed.
t The terminology in this example derives from projective geometry, however,
Exercise 2 shows that 2 can more aptly be described as a twisted sphere.
134 CALCULUS ON A SURFACE [Chap. IV
8.3 Definition Let a: I — > M be a curve in an abstract surface M. For
each t in /, the velocity vector a (t) is the function such that
«'(0W = ^ (0
for every differentiable realvalued function / on M.
Thus a (t) is a realvalued function whose domain is the set of all differ
entiable functions on M. This is all we need to generalize the calculus for
surfaces in E to the case of an abstract surface.
The reader may feel he has gone far enough in the direction of abstrac
tion, but in one more step we shall have gone all the way.
We now have a calculus for E" (Chapter I) and a calculus for surfaces.
These are strictly analogous, but analogies in mathematics (although use
ful at first) are, in the long run, annoying. What we need is a single calculus,
of which these two will be special cases. The most general object on which
calculus can be conducted is called a manifold. It is simply an abstract
surface of arbitrary dimension n.
8.4 Definition An ndimensional manifold M is a set furnished with a
collection (P of abstract patches (onetoone functions x: D — ► M, D an
open set in E n ) such that
(1 ) M is covered by the images of the patches in the collection (P.
(2) For any two patches x, y in the collection (P, the composite func
tions y _1 x and x~*y are Euclideandifferentiable (and defined on open sets
inE n ).
Thus a surface (Definition 8.1) is the same thing as a twodimensional
manifold. The Euclidean space E" is a very special ndimensional manifold;
its patch collection consists only of the identity function.
To keep this definition as close as possible to that of a surface in E 3 , we
have deviated slightly from the standard definition of manifold in which it
is usually the inverse functions x _1 : x (D ) — » D that are axiomatized.
The calculus of an arbitrary ndimensional manifold M is defined in the
same way as in the special case, n = 2, of an abstract surface. Differentiable
functions, tangent vectors, vector fields, and mappings are gotten exactly
as before: We need only replace i = 1, 2 by i = 1, 2, • • • , n. Differential
forms on a manifold M have the same general properties as in the case
n = 2, which we have explored in Sections 4, 5, and 6. But there are p
forms for ^ p ^ n, so when the dimension n of M is large, the situation
becomes rather more complicated than for n = 2, and more sophisticated
techniques are called for.
Whenever calculus appears in mathematics and its applications, mani
folds will also be found, and higher dimensional manifolds turn out to be
Sec. 8]
MANIFOLDS
185
FIG. 4.50
important in problems (both pure and applied) that initially seem to in
volve only dimensions 2 or 3. For example, we now describe a fourdimen
sional manifold that has already appeared, implicitly at least, in this
chapter.
8.5 Example The tangent bundle of a surface. If M is a surface, let T (M )
be the set of all tangent vectors to M at all points of M . (For the sake of
concreteness we shall think of M as a surface in E 3 , but M could just as well
be an abstract surface, or indeed a manifold of any dimension.) Now M
itself has dimension 2 and each tangent plane T P (M) has dimension 2, so
T(M) will turn out to have dimension 4. To get the patch collection (P
that will make the set T(M) a manifold, we shall derive from each patch
xinla patch x in T(M). Given x: D — > M, let 6 be the open set in E 4
consisting of all point (p h p 2 , p 3 , p 4 ) for which (p u p 2 ) is in D. Then let
x: D — * T(M) be the function such that
Xfal, P2, PZ, Pi) = P»X«(pi, Pi) + P4X B (P1, Pi).
(In Fig. 4.50 we identify E 2 with the x x Xi plane of E 4 and deal as best we
can with dimension 4.)
Using Exercise 3 of Section 3 and the proof of Lemma 3.6, it is not diffi
cult to check that (1 ) each such function x is onetoone, hence is a patch
in T{M), in the sense of Definition 8.4, and (2) the collection & of all
such patches satisfies the two conditions in Definition 8.4. Thus T(M) is
a fourdimensional manifold called the tangent bundle of M.
EXERCISES
Show that a surface M is nonorientable if there is a closed curve
a: [0, 1] — * M and a vector field Y on a such that
(a) Y and a are linearly independent at each point.
(b) F(l) = F(0).
186 CALCULUS ON A SURFACE [Chap. IV
2. Establish the following properties of the projective plane 2:
(a) If P: 2 — ■» 2 is the projection, then each tangent vector to 2
is the image under P* of exactly two tangent vectors to 2 — of
the form v p and ( — v)_„.
(b) 2 is compact, connected, and nonorientable.
(Hint: For (a), use Ex. 5 of IV.5.) Although the proof is difficult,
every compact surface in E is orientable — thus 2 is not diffeomorphic to
any surface in E 3 .
3. Prove that the tangent bundle (8.5) is a manifold. (If x'and y are
overlapping patches in M, find an explicit formula for y 1 x.)
4. If M is the image of a single patch x: E — > M, show that the tangent
bundle T(M) is diffeomorphic to E 4 .
5. (Plane with two origins). Let M consist of all ordered pairs of real
numbers (u, v) and one additional point 0*. Let x and y be the func
tions from E 2 to M such that
x(u, v) = y(u, v) = (u, v) if (u, v) ^ (0, 0),
but
x (0, 0) = = (0, 0) and y(0, 0) = 0*.
Prove that:
(a) The abstract patches x and y make M a surface.
(b) M is connected.
(c) The function F: M — » M is a mapping, where F(0) = 0* and
F(0*) = 0, but F(p) = p for all other points of M.
Surfaces such as this one are troublesome to deal with; we eliminate
them by adding an additional hypothesis to Definition 8.1: For any
points p 9^ q of M there exist abstract patches x and y in (P such that
p is in x(D), q is in y (E), and x(D) and y (E) do not meet (Hausdorff
axiom).
6. Let V be a vector field on a surface M. A curve a in M is an integral
curve of V provided a (t) = V(a(t)) for all t. Thus an integral curve
has at each point the velocity prescribed by V. If a(0) = p, we say
that a starts at p.
(a) In the special case M = E 2 , show that the curve t — > (ai(t), a^(t))
is an integral curve of v = /i C/i + U Ui starting at (a, b) if
and only if
tj = /i(«i» ai) fai(0) = a
and
77 = Mai, a 2 ) la 2 (0) = b
at
Sec. 9] SUMMARY 187
The theory of differential equations predicts a unique solution for
such systems.
(b) Find the integral curve of
v = —u 2 Ui + uvUz on E 2
which starts at the point (1, —1). {Hint: The differential equa
tions in this case can be solved by elementary methods. Use the
arbitrary constants in the solution to make the starting point
(1, DO
7. Prove that every vector field V on a surface M has an integral
curve starting at any given point. (Hint: Pick a patch x in M, with
x(a,b) = p, and let v be the vector field on E 2 such that x* (v) = V. )
8. Prove that every surface of revolution is diffeomorphic to either a
torus or a circular cylinder. (Similarly an augmented surface of revolu
tion — Ex. 12 of IV. 1 — is diffeomorphic to either a plane or a sphere.)
9. (Cartesian products). If M and N are surfaces, let M X N be the set
of all ordered pairs (p, q), with p in M and q in N. If x: D — ► M and
y: E — > N are patches, let D X E be the region in E 4 consisting of
all points (u, v, Mi, Vi) with (v, v) in D, (u x , vi) in E. Then define
x X y: D X E ^ M X N by
(x X y)(u, v, Mi, vi) = (x(u,v),y(ui,Vi)).
Prove that the collection (P of all such abstract patches makes M X N
a manifold (of dimension 4). M X AT is called the Cartesian product
of M and N.
The same scheme works for any two manifolds — for example,
E 1 X E 1 is precisely E 2 .
10. If M is an abstract surface, a proper imbedding of M in E is a oneto
one regular mapping F: M — > E such that the inverse function
F~ x : F (M) — * M is continuous. Prove that the image F (M) of a proper
imbedding is a surface in E 3 (Definition 1.2) and is diffeomorphic to M.
If F: M — > E 3 is merely regular, then F is an immersion of M in E
and the image F(M) is sometimes called an "immersed surface," even
though it need not satisfy Definition 1.2.
9 Summary
In this chapter we have progressed from the familiar notion of surface in
E to the general notion of manifold. Let us now reverse this process: An
ndimensional manifold M is a space that — near each point — is like the
Euclidean space E n . Every manifold has a calculus consisting of differentia
188 CALCULUS ON A SURFACE [Chap. IV
ble functions, tangent vectors, vector fields, mappings, and, above all,
differential forms. The simplest manifold of dimension n is E n itself. A
twodimensional manifold is called a surface. Some surfaces appear in E 3 ,
some do not. This theory all comes from the usual elementary calculus on
the bestknown manifold of all, the real line. But the calculus of every
manifold behaves in the same general way.
CHAPTER
V
Shape Operators
In Chapter II we measured the shape of a curve in E 3 by its curvature and
torsion functions. Now we consider the analogous measurement problem
for surfaces. It turns out that the shape of a surface M in E 3 is described
infinitesimally by a certain linear operator £ denned on each of the tangent
planes of M . As with curves, to say that two surfaces in E 3 have the same
shape means simply that they are congruent. And just as with curves, we
shall justify our infinitesimal measurements by proving that two surfaces
with "the same" shape operators are, in fact, congruent. The algebraic
invariants (determinant, trace, • • • ) of its shape operators thus have
geometric meaning for the surface M. We shall investigate this matter in
detail and find efficient ways to compute these invariants, which we test
on a number of geometrically interesting surfaces.
From now on the notation M <Z E 3 means a connected surface M in
E 3 as defined in Chapter IV.
1 The Shape Operator ofMc E 3
Suppose that Z is a Euclidean vector field (Definition 3.7 of Chapter IV) on
a surface M in E 3 . Although Z is defined only at points of M, the covariant
derivative V V Z (Chapter II, Section 5) still makes sense as long as v is
tangent to M. As usual, V V Z is the rate of change of Z in the v direction,
and there are two main ways to compute it.
Method 1. Let a be a curve in M that has initial velocity a (0) = v.
Let Z a be the restriction of Z to a, that is, the vector field t — > Z(a (t) ) on
a (Fig. 5.1). Then
189
190
SHAPE OPERATORS
[Chap. V
V r Z
FIG. 5.1
V„Z = (Z a )'(0)
where the derivative is that of Chapter II, Section 2.
Method 2. Express Z in terms of the natural frame field of E 3 by
z = E Zi Ui.
Then
v„z = EvNC/i
where the directional derivative is that of Chapter IV, Section 3.
It is easy to check that these two methods give consistent results. Note
that even if Z is a tangent vector field, the co variant derivative V„Z need
not be tangent to M.
It follows immediately from Theorem 7.5 of Chapter IV that if M is an
orientable surface in E 3 , then there is a unit normal vector field U on M.
In fact, if Z is a nonvanishing normal vector field, then U = Z/ \\ Z \\ is
still normal, and has unit length. Since M is now assumed to be connected,
there are exactly two unit normal vector fields U and — U defined on the whole
surface M. But even when M is not orientable, unit normals U and — U
are still available on some neighborhood of each point of p of M (see Exer
cises for Chapter IV, Section 7 ) .
We are now in a position to find a mathematical measurement of the
shape of a surface in E .
1.1 Definition If p is a point of M, then for each tangent vector v to M
at p, let
£ p (v) = V V U
where U is a unit normal vector field on a neighborhood of p in M. S p is
called the shape operator of M at p (derived from C/).f (Fig. 5.2.)
t The minus sign artificially introduced in this definition will sharply reduce the
total number of minus signs needed later on.
See. 1] THE SHAPE OPERATOR OF M C E 3 191
Ufa)
FIG. 5.2
The tangent plane of M at any point q consists of all Euclidean vectors
orthogonal to C/(q). Thus the rate of change V„J7 of U in the v direction
tells how the tangent planes of M are varying in the v direction — and
this gives an infinitesimal description of the way M itself is curving in E 3 .
Note that if U is replaced by — U, then S p changes to — S p .
1.2 Lemma For each point p of M CI E 3 , the shape operator is a linear
operator
S p : T P (M) + T P (M)
on the tangent plane of M at p.
Proof. In Definition 1.1, U is a unit vector field, so U'U = 1. Thus by
a Leibnizian property of covariant derivatives,
= y[U*U\ = 2V V UU(p) = 2£ p (v).tf(p)
where v is tangent to M at p. Since U is also a normal vector field, it follows
that S p (v) is tangent to M at p. Thus S P is a function from T P (M) to
T P (M). (It is to emphasize this that we use the term "operator" instead of
' 'transformation. ' ' )
The linearity of S p is a consequence of a linearity property of covariant
derivatives.
S p (av + 6w) = —Vav+bwU =  (aV v U + bVJU)
= aS p (y) + bS p (w). 
At each point p of M a E there are actually two shape operators ±S P
derived from the two unit normals ±U near p. We shall refer to all of
these, collectively, as the shape operator S of M. Thus if a choice of unit
normal is not specified, there is a (relatively harmless) ambiguity of sign.
1.3 Example Shape operators of some surfaces in E 3 . (1) Let S be
the sphere of radius r consisting of all points p of E 3 with  p  = r. Let
U be the "outward normal" on S. Now as U moves away from any point
p in the direction v, evidently U topples forward in the exact direction of
v itself (Fig. 5.3). Thus S(\) must have the form — cv.
In fact, using gradients as in Example 3.9 of Chapter IV, we find
192
SHAPE OPERATORS
[Chap. V
C/ = iZ
But then
V V U =EvW^(p) =
Thus S(\) = — v/r for all v. So the shape
operator S is merely scalar multiplication by
— 1/r. This uniformity in S reflects the round
fig. 5.3 ness of spheres: They bend the same way in all
directions at all points.
(2) Let P be a plane in E 3 . A unit normal vector field U on P is evidently
parallel in E (constant Euclidean coordinates) (Fig. 5.4). Hence
S(v) = V V U =
for all tangent vectors v to P. Thus the shape operator is identically zero
— to be expected, since planes do not bend at all.
(3) Let C be the circular cylinder x + y 2 = r 2 in E 3 . At any point p
of C, let ei and e 2 be unit tangent vectors, with e x tangent to the ruling of
the cylinder through p, and ^ tangent to the crosssectional circle.
Use the outward normal U as indicated in Fig. 5.5.
Now, when U moves from p in the ei direction, it stays parallel to itself
just as on a plane; hence *S(d) = 0. When U moves in the e 2 direction, it
topples forward exactly as on a sphere of radius r; hence *S(e 2 ) = — e 2 /r.
In this way £ describes the "half flat, halfround" shape of a cylinder.
(4) The saddle surface M: z = xy. For the moment we investigate S
only at p = (0, 0, 0) in M . Since the x and y axes of E 3 lie in M , the vectors
iii = (1,0,0) and u 2 = (0, 1,0) are tangent to M at p. We use the "up
ward" unit normal U, which at p is (0,0, 1). Along the x axis, U stays
U
U(p)
FIG. 5.4
FIG. 5.5
Sec. 1]
THE SHAPE OPERATOR OFMCP
193
FIG. 5.6
orthogonal to the x axis, and as we proceed in the ui direction, U swings
from left to right (Fig. 5.6). In fact, a routine computation (Exercise 3)
shows that V Ul U = u 2 . Similarly we find V U2 U = — u x .
Thus the shape operator of M at p is given by the formula
S(aui + 6u 2 ) = b\ii + au 2 .
These examples clarify the analogy between the shape operator of a
surface and the curvature and torsion of a curve. In the case of a curve,
there is only one direction to move, and k and t measure the rate of change
of the unit vector fields T and B (hence N). For a surface only one unit
vector field is intrinsically determined — the unit normal U. Furthermore,
at each point, there are now a whole plane of directions in which U can
move, so that rates of change of U are measured, not numerically, but by
the linear operators S.
1.4 Lemma For each point p of I c E 3 the shape operator
S: T P M > T P M
is a symmetric linear operator; that is,
/S(v)»w = S(w)'\
for any pair of tangent vectors to M at p.
We postpone the proof of this crucial fact to Section 4, where it occurs
naturally in the course of general computations.
From the viewpoint of linear algebra, a symmetric linear operator on a
twodimensional vector space is a very simple object indeed. For a shape
operator, its characteristic values and vectors, its trace and determinant,
all turn out to have geometric meaning of first importance for the surface
M a E 3 .
194 SHAPE OPERATORS [Chap. V
EXERCISES
1 . Let a be a curve in M (Z E 3 . If U is a unit normal of M restricted to
the curve a, show that S(a') = — V .
2. Consider the surface M: z = f(x,y), where
/(0,0) = /,(0,0) = /,(0,0) = 0.
(The subscripts indicate partial derivatives.) Show that
(a) The vectors ui = Ui(0) and u 2 = ^(O) are tangent to M at the
origin 0, and
~ UUl  fyU2 + Us
Vi + U + /„»
is a unit normal vector field on M.
(b) £( Ul ) =/«(0,0)u 1 +/x„(0,0)ua
^(U2) =/,x(0,0) Ul +/ yw (0,0)u 2 .
(Note: The square root in the denominator is no real problem here be
cause of the special character of/ at (0, 0). In general, direct computa
tion of S is difficult, and in Section 4 we shall establish indirect ways of
getting at it.)
3. (Continuation) . In each case, express S(aui + 6u 2 ) in terms of ui and
u 2 , and determine the rank of *S at (rank S is dimension of image S:
0,1, or 2).
(a) z = xy. (c) z = (x + yf.
(b) z = 2x 2 + y. (d) z = xy.
4. Let M be a surface in E oriented by a unit normal vector field
U = giUi + QiVi + gzUz.
Then the Gauss mapping G: M —* 2 of M sends each point p to the
point (<7i(p), <72(p)> <73 (p)) of the unit sphere 2. Pictorially: Move
U(p) to the origin by parallel motion; there it points to G(p) (Fig. 5.7).
U(p)
FIG. 5.7
Sec. 2] NORMAL CURVATURE 195
Thus G completely describes the turning of U as it traverses M.
For each of the following surfaces, describe the image G(M) of the
Gauss mapping in the sphere 2 (use either normal) :
(a) Cylinder, x + y 2 = r .
(b) Cone, z = y/x* + y 2 .
(c) Plane, x + y + z = 0.
(d) Sphere, (x  l) 2 + y 2 + (z + 2) 2 = 1.
5. Let G: T — > 2 be the Gauss mapping of the torus T (as in IV. 2.6)
derived from its outward unit normal U. What are the image curves
under G of the meridians and parallels of T? Which points of 2 are the
image of exactly two points of T?
6. Let G: M — > 2 be the Gauss mapping of the saddle surface M : z = xy
derived from the unit normal U obtained as in Exercise 2. What is the
image under G of one of the straight lines, y constant, in M? How much
of the sphere is covered by the entire image G(M) ?
7. Show that the shape operator of M is (minus) the derivative of its
Gauss mapping: If S and G: M > 2 both derive from U, then S(\)
and — (j* (v) are parallel for every tangent vector v to M.
8. An orientable surface has two Gauss mappings derived from its two
unit normals. Show that they differ only by the antipodal mapping of
2 (Ex. 5 of IV. 5). Define a Gausstype mapping for a nonorientable
surface in E .
9. If V is a tangent vector field on M (with unit normal U), then by the
pointwise principle, S(V) is the tangent vector field on M whose value
at each point p is S p (V(p)). Show that
S(V)W = VyWU.
Deduce that the symmetry of S is equivalent to the assertion that
the bracket
[V, W] = v v w  v w v
of two tangent vector fields is again a tangent vector field.
2 Normal Curvature
Throughout this section we shall work in a region of M CI E which has been
oriented by the choice of a unit normal vector field U, and we use the shape
operator S derived from U.
The shape of a surface in E 3 influences the shape of the curves in M.
196
SHAPE OPERATORS
[Chap. V
FIG. 5.8
2.1 Lemma If a is a curve inMc E 3 , then
a ".U = S(a')a.
Proof. Since a is in M, its velocity a is always tangent to M . Thus
a • U = 0, where as in Section 1 we restrict U to the curve a. Differentiation
yields
<*".[/ + d'Xf = 0.
But from Section 1, we know that S(a ) = — U. Hence
a ».U = if. a ' = S(d).d. 
Geometric interpretation: at each point a" ' *U is the component of accel
eration a" normal to the surface M (Fig. 5.8). The lemma shows that this
component depends only on the velocity a and the shape operator of M.
Thus all curves in M with a given velocity v at point p will have the same nor
mal component of acceleration at p, namely, $(v)«v. This is the component
of acceleration which the bending ofMinJZ forces them to have.
Thus if we standardize v by reducing it to a unit vector u, we get a
measurement of the way M is bent in the u direction.
2.2 Definition Let u be a unit vector tangent toMcE at a point p.
Then the number k(u) = $(u)»uis called the normal curvature of M in
the u direction.
To make the term direction precise, we define a tangent direction to M at
p to be a onedimensional subspace L of T P (M), that is, a line through the
zero vector (located for intuitive purposes at p) (Fig. 5.9). Any nonzero
tangent vector at p determines a direction L, but we prefer to use one of
FIG. 5.9
Sec. 2]
NORMAL CURVATURE
197
FIG. 5.10
the two unit vectors ±u in L. Note that
fc(u) = S(u)u = S(u).(u) = fc(u).
Thus, although we evaluate k on unit vectors, it is, in effect, a realvalued
function on the set of all tangent directions to M.
Given a unit tangent vector to M at p, let a be a (unit speed) curve in
M with initial velocity a'(0) = u. Using the Frenet apparatus of a, the
preceding lemma gives
fc(u) = S(u)u = a"(0).£/(p) = K(0)iV(0).t/(p)
= k(0) cost?.
Thus the normal curvature of M in the u direction is k(0) cos #, where
k(0) is the curvature of a at a (0) = p, and # is the angle between the
principal normal N(0) and the surface normal U(p), as in Fig. 5.10.
Given u, there is a natural way to choose the curve so that # is or v.
In fact, if P is the plane determined by u and C/(p), then P cuts from M
(near p) a curve <r called the normal section of M in the u direction. If we
give a unitspeed parametrization with <r (0) = u, it is easy to see that
N(0) = dbU(p). (er"(0) = k(0)N(0) is orthogonal to </(0) = u and
tangent to P.) Thus for a normal section in the u direction (Fig. 5.11),
fc(u) = K a (0)N(0)U(p) = ±k„(0).
Thus it is possible to make a reasonable estimate of the normal curva
tures in various directions on a surface M CI E 3 by picturing what the cor
responding normal sections would look like. We know that the principal
normal N of a curve tells in which direction it is turning. Thus the preceding
discussion gives geometric meaning to the sign of the normal curvature
k(u) (relative to our fixed choice of U).
198
SHAPE OPERATORS
[Chap. V
FIG. 5.11
(1) If fc(u) > 0, then AT (0) = U(p), so the normal section a is bending
toward U(p) at p (Fig. 5.12). Thus in the u direction the surface M is
bending toward U(p).
U(p)
N(0) = CT(p)
FIG. 5.12
N(0)
FIG. 5.13
(2) If k(u) < 0, then JV(0) = — U(p), so the normal section a is
bending away from U(p) at p. Thus in the u direction M is bending away
from U(p) (Fig. 5.13).
(3) If k(u) = 0, then k,(0) = (iV(0) is undefined). Here the normal
section a is not turning at <r(0) = p. We cannot conclude that in the u
direction M is not bending at all, since k might be zero only at o(O) = p.
But we can conclude that its rate of bending is unusually small.
In different directions at a fixed point p, the surface may bend in quite
different ways. For example, consider the saddle surface z = xy in Example
1.3. If we identify the tangent plane of M at p = (0, 0, 0) with the xy plane
of E 3 , then clearly the normal curvature in the direction of the x and y axes
is zero, since the normal sections are straight lines. However, Fig. 5.6 shows
that in the tangent direction given by the line y = a;, the normal curva
ture is positive, for the normal section is a parabola bending upward.
(C/(p) = (0,0,1) is "upward.") But in the direction of the line y = —x,
normal curvature is negative, since this parabola bends downward.
Sec. 2]
NORMAL CURVATURE
U(p)
199
E*
(^j2^^«
T,(M\
'•* •■•''■* **■•?& V
FIG. 5.14
Let us now fix a point p of M C E 3 and imagine that a unit tangent
vector u at p revolves, sweeping out the unit circle in the tangent plane
T P (M). From the corresponding normal sections, we get a moving picture
of the way M is bending in every direction at p (Fig. 5.14).
2.3 Definition Let p be a point of M CZ E 3 . The maximum and minimum
values of the normal curvature A;(u) of M at p are called the principal
curvatures of M at p, and are denoted by fa and fa. The directions in which
these extreme values occur are called principal directions of M at p. Unit
vectors in these directions are called principal vectors of M at p.
Using the normalsection scheme discussed above, it is often fairly easy
to pick out the directions of maximum and minimum bending. For example,
if we use the outward normal (U) on a circular cylinder C, then the normal
sections of all bend away from U, so k (u) ^ 0. Furthermore, it is reasonably
clear that the maximum value fa = occurs only in the direction of a rul
ing, minimum value fa < occurs only in the direction tangent to a cross
section, as in Fig. 5.15.
FIG. 5.15
200 SHAPE OPERATORS [Chap. V
An interesting special case occurs at points p for which fa = fa. The
maximum and minimum normal curvature being equal, it follows that k (u)
is constant: M bends the same amount in all directions at p (and all direc
tions are principal).
2.4 Definition A point p of M C E 3 is umbilic provided the normal
curvature A;(u) is constant on all unit tangent vectors u at p.
For example, what we found in (1 ) of Example 1.3 was that every point
of the sphere 2 is umbilic, with fa = fa = — 1/r.
2.5 Theorem (1) If p is an umbilic point of M C E 3 , then the shape
operator S at p is just scalar multiplication by k = fa = fa.
(2 ) If p is a nonumbilic point, fa ?* fa, then there are exactly two prin
cipal directions, and these are orthogonal. Furthermore, if d and e2 are
principal vectors in these directions, then
S(ei) = fae! /S(e 2 ) = A^d
In short, the principal curvatures of M at p are the characteristic values
of S, and the principal vectors of M at p are the characteristic vectors of S.
Proof. Suppose that k takes on its maximum value fa at d, so
fa = k(ei) = Sid) »ei.
Let C2 be merely a unit tangent vector orthogonal to ei (presently we shall
show that it is also a principal vector. )
If u is any unit tangent vector at p, we write
u = u(#) = cei + se2
where c = cos ■&, s = sin & (Fig. 5.16). Thus normal curvature k at p be
comes a function on the real line: k(&) = k(u(&)).
For 1 ^ i, j <^ 2, let Sa be the number S(ei)*ej. Note that S n = fa, and
by the symmetry of the shape operator, $ 12 = S^. We compute
&(#) = »S(cei + se 2 )«(cei + se 2 )
(1)
= cS n + 2scSu + s 2 £ 22 .
Hence
^ (*) = 2 SC (£ 22  Sn) + 2(c 2  s 2 )S 12 . (2)
d&
If # = 0, then c = 1 and s = 0, so u(0) = d. Thus, by assumption,
fc(#) is a maximum at & = 0, so (dk/d&)(0) = 0. It follows immediately
from (2) that S n = 0.
Since d, e 2 is an orthonormal basis for T P (M), we deduce by orthonormal
Sec. 2]
NORMAL CURVATURE
201
expansion that
£(ei) = S u ei /S(e 2 ) = &2e 2 .
(3)
Now if p is umbilic, then S 22 = fcfe) is the same as S n = &(ei) = h, so
(3) shows that S is scalar multiplication by h = k 2 .
If p is not umbilic, we look back at (1), which has become
fc(#) = c % + s 2 &2. (4)
Since h is the maximum value of fc(#), and fc(t^) is now nonconstant, it
follows that h > S*. But then (4) shows: (a) the maximum value h is
taken on only when c = ±1, s = 0, that is, in the d direction, and (b) the
minimum value k 2 is £22, and is taken on only when c = 0, s = ±1, that is,
in the e 2 direction. This proves the second assertion in the theorem, since
(3 ) now reads :
£(ei) = hd, SM = k^. I
Contained in the preceding proof is Euler's formula for the normal curva
ture of M in all directions at p.
2.6 Corollary Let k u k 2 and e u d be the principal curvatures and vectors
oflcE 8 at p. Then if u = cos tfei + sin t?e 2 , the normal curvature of
M in the u direction is (Fig. 5.16)
k(u) = h cos 2 # + kv sin 2 t?.
Here is another way to show how the principal curvatures h and fc 2
control the shape of M near an arbitrary point p. Since the position of M
in E 3 is of no importance, we can assume that (1) p is at the origin of E ,
(2) the tangent plane f p (M) is the xy plane of E 3 , and (3) the x and y axes
are the principal directions. Near p, M can be expressed as M: z = f(x,y),
as shown in Fig. 5.17, and the idea is to construct an approximation of M
near p by using only terms up to quadratic in the Taylor expansion of the
(x,y,f(x,y)
T P (M)
FIG. 5.16
FIG. 5.17
202 SHAPE OPERATORS [Chap. V
function/. Now (1) and (2) imply/ = f x = f y ° = 0, where the subscripts
indicate partial derivatives, and the superscript zero denotes evaluation at
x = 0, y = 0. Thus the quadratic approximation of / near (0, 0) reduces to
f(x,y) ~ h(f xx x + 2f xy xy + f yy y 2 ).
In Exercise 2 of Section 1 we found that for the tangent vectors
ux = (1,0,0) and u 2 = (0,1,0)
at p =
£(ui) = V Ml C7 = f xx jx 1 +/° J/ u 2
S(U 2 ) = —V U2 U= f xy U! + flyU 2 .
By condition (3) above, ui and u 2 are principal vectors, so it follows
from Theorem 2.5 that h = f xx , k 2 = f yy , and /J, = 0.
Substituting these values in the quadratic approximation of /, we con
clude that the shape of M near p is approximately the same as that of the
surface
M:z = \{hx 2 + k 2 y 2 )
near 0. M is called the quadratic approximation of M near p. It is an ana
logue for surfaces of a Frenet approximation of a curve.
From Definition 2.2 through Corollary 2.6 we have been concerned with
the geometry of M C E 3 near one of its points p. These results thus apply
simultaneously to all the points of the oriented region on which, by our
initial assumption, the unit normal U is defined. In particular then, we
have actually defined principal curvature functions ki and k 2 on 0. At each
point p of 0, &i(p) and A; 2 (p) are the principal curvatures of M at p. We
emphasize that these functions are only defined "modulo sign": If U is
replaced by —U, they become — h and — k 2 .
EXERCISES
1. Use the results of Example 1.3 to find the principal curvatures and
principal vectors of
(a) The cylinder, at every point.
(b) The saddle surface, at the origin.
2. If v is a nonzero tangent vector (not necessarily of unit length), show
that the normal curvature of M in the direction determined by v is
k(v) = S(\)'\/y\.
3. For each integer n ^ 2, let a n be the curve t — > (r cos t, r sin t, ±t n )
Sec. 3] GAUSSIAN CURVATURE 203
in the cylinder M: x + y 2 = r 2 . These curves all have the same velocity
at t = 0; test Lemma 2.1 by showing that they all have the same normal
component of acceleration at t = 0.
4. For each of the following surfaces, find the quadratic approximation
near the origin:
(a) z = exp(z 2 + y 2 )  1.
(b) z = log cos x — log cos y.
(c) z = (x + Zy)\
5. Justify the first sentence in the proof of Theorem 2.5 : Show that k has a
maximum value.
3 Gaussian Curvature
In the preceding section we found the geometrical meaning of the charac
teristic values and vectors of the shape operator. Now we examine the de
terminant and trace of S.
3.1 Definition The Gaussian curvature ofMcE is the real valued func
tion K = det S on M. Explicitly, for each point p of M, the Gaussian curva
ture K(p) of M at p is the determinant of the shape operator S of M at p.
The mean curvature of M c E 3 is the function H = \ trace S. Gaussian
and mean curvature are expressed in terms of principal curvature by
3.2 Lemma K = kfa, H = (h + k 2 )/2.
Proof. The determinant (and trace) of a linear operator may be defined
as the common value of the determinant (and trace) of all its matrices.
If e x and 62 are principal vectors at a point p, then by Theorem 2.5, we have
>S(ei) = fci(p)ei and /S(e 2 ) = fc 2 (p)e2. Thus the matrix of S at p with re
spect to d, e 2 is
'fcx(p)
^0 fc 2 (p),
This immediately gives the required result. 
A significant fact about the Gaussian curvature: It is independent of the
choice of the unit normal U. If U is changed to — U, then the signs of both
ki and fc 2 change, so K = kik 2 is unaffected. This is obviously not the case
with mean curvature H = (h { k 2 )/2, which has the same ambiguity of
sign as the principal curvatures themselves.
The normal section method in Section 2 lets us tell, by inspection, ap
proximately what the principal curvatures of M are at each point. Thus we
204
SHAPE OPERATORS
[Chap. V
U(p)
hip) <
hip) <
U(p)
FIG. 5.18
FIG. 5.19
get a reasonable idea of what the Gaussian curvature K = hh is at each
point p by merely looking at the surface M. In particular, we can usually
tell what the sign of Kip) is — and this sign has an important geometric
meaning, which we now illustrate.
3.3 Remark The sign of Gaussian curvature at a point p.
(1) Positive. If Kip) > 0, then by Lemma 3.2, the principal curvatures
hip) and A; 2 (p) have the same sign. By Corollary 2.6, either k (u) > for
all unit vectors u at p or k(u) < 0. Thus M is bending away from its
tangent plane T P (M) in all tangent directions at p. (Fig. 5.18).
The quadratic approximation of M near p is the paraboloid
2z = hip)x 2 + hip)y\
(2) Negative. If Kip) < 0, then by Lemma 3.2 the principal curvatures
h (p) and k 2 (p) have opposite signs. Thus the quadratic approximation of
M near p is a hyperboloid, so M also is saddleshaped near p (Fig. 5.19).
(3) Zero. If K(p) = 0, then by Lemma 3.2 there are two cases:
(a) Only one principal curvature is zero, say
h(p) * 0, A; 2 (p) = 0.
(b) Both principal curvatures are zero:
hip) = fc 2 ( P ) = 0.
In case (a) the quadratic approximation is the cylinder 2z = hip)x 2 , so
M is troughshaped near p (Fig. 5.20).
In case (b), the quadratic approximation reduces simply to the plane
z = 0, so we get no information about the shape of M near p.
A torus of revolution T provides a good example of these different cases.
At points on the outer half of T, the torus bends away from its tangent
plane as one can see from Fig. 5.21 ; hence K > on 0. But near each point
p of the inner half d, T is saddleshaped and cuts through T P iM). Hence
K < on d.
Sec. 3]
GAUSSIAN CURVATURE
205
U(P)
*i(p) >
M l hip) =
FIG. 5.20
FIG. 5.22
Near each point on the two circles (top and bottom) which separate
and #, the torus is troughshaped; hence K = there. (A quantitative
check of these qualitative results is given in Section 6.)
In case 3 (b ) above, where both principal curvatures vanish, p is called a
planar point of M . (There are no planar points on the torus.) For example,
the central point p of a monkey saddle, say
M: z = x(x + VSy) (x  VSy),
is planar. Here three hills and valleys meet, as shown in Fig. 5.22. Thus
p must be a planar point — the shape of M near p is too complicated for the
other three possibilities in Remark 3.3.
We consider now some ways to compute Gaussian and mean curvature.
3.4 Lemma If v and w are linearly independent tangent vectors at a
point p of M cz E 3 , then
S(v) X S(w) = K(p)v X w
S(v) X w + v X S(w) = 2H(p)v X w.
Proof. Since v, w is a basis for the tangent plane T P (M), we can write
iS(v) = av + 6w
S(w) = cv + dw.
Thus
206
SHAPE OPERATORS
[Chap. V
a b
c d
is the matrix of S with respect to the basis v, w. Hence
K(p) = det S = ad  be H(p) = \ trace S = (o + d).
Using standard properties of the cross product, we compute
S(y) X S(w) = (ax + 6w) X (cv + dw)
= (ad  be) v X w = K(p) v X w
and a similar calculation gives the formula for H(p).
I
Thus if V and W are tangent vector fields that are linearly independent
at each point of an oriented region, we have vector field equations
S(V) X S(W) = KV XW
S(V) XW +V X S(W) =2HV XW.
These may be solved for K and H by dotting each side with the normal
vector field V XW, and using the Lagrange identity (Exercise 6 ) . We then
find
K =
SVV SVW
SWV SWW
H =
V*V V'W
WV W'W
SVV sv
WV w
•W
W
+
V'V V'W
SWV SWW
2
W
V
V
V'W
W'W
(The denominators are never zero, since the independence of V and W is
equivalent to (V X W)'(V X W) > 0.) In particular the functions K
and H are differ entiable.
Once K and H are known, it is a simple matter to find ki and k 2 .
3.5 Corollary On an oriented region in M, the principal curvature
functions are
k h k 2 = H ± VH 2  K.
Proof. To verify the formula it suffices to substitute
K = A;^ and H = (h + A*)/2,
and note that
H 2  K =
(a* + k 2 y
— kik 2
(fa — faf
Sec. 3] GAUSSIAN CURVATURE 207
A more enlightening derivation (Exercise 4) uses the characteristic poly
nomial of S.
This formula shows only that ki and k 2 are continuous functions on 0;
they need not be differentiable since the squareroot function is badly be
haved at zero. The identity in the proof shows that H 2 — K is zero only at
umbilic points, however, so ki and k 2 are differentiable on any oriented region
free of umbilics.
A natural way to single out special types of surfaces in E 3 is by restric
tions on Gaussian and mean curvature.
3.6 Definition A surface M in E 3 is flat provided its Gaussian curvature
is zero, and minimal provided its mean curvature is zero.
As expected, a plane is flat, for by Example 1.3 its shape operators are
all zero, so K = det S = O.\0n a circular cylinder, (3 ) of Example 1.3 shows
that S is singular at each point p, that is, has rank less than the dimension
of the tangent plane T P (M). Thus, although S itself is never zero, its de
terminant is always zero, so cylinders are also flat. This terminology seems
odd at first for a surface so obviously curved, but it will be amply justified
in later work.
Note that minimal surfaces have Gaussian curvature K ^ 0, because
if H = (fci + k 2 )/2 = 0, then h = k 2 , so K = hk 2 ^ 0.
Another notable class of surfaces consists of those with constant Gaussian
curvature. As mentioned earlier, Example 1.3 shows that a sphere of radius
r has ki = k 2 = —1/r (for U outward). Thus the sphere 2 has constant
positive curvature K = 1/r 2 : The smaller the sphere, the larger its curvature.
We shall find many examples of these various special types of surface
as we proceed through this chapter.
EXERCISES
1. Show that there are no umbilics on a surface with K < 0, and that if
K ^ 0, umbilic points are planar.
2. Let iii and u 2 be orthonormal tangent vectors at a point p of M. What
geometric information can be deduced from each of the following condi
tions on S at p?
(a) 5(u!).u s = 0. (c) S( Ul ) X #(u 2 ) = 0.
(b) 5(u0 + S(u 2 ) = 0. (d) S( Ul )S(u 2 ) = 0.
3. (Mean curvature). Prove that
(a) the average value of the normal curvature in any two orthogonal
directions at p is H(p). (The analogue for K is false.)
208 SHAPE OPERATORS [Chap. V
(b) H(p) = (1/2*) f **(*) dfi,
where k(&) is the normal curvature, as in Corollary 2.6.
4. The characteristic polynomial of an arbitrary linear operator S is
p(k) = det(A  kl),
where A is any matrix of S.
(a) Show that the characteristic polynomial of the shape operator is
k 2  2Hk + K.
(b) Every linear operator satisfies its characteristic equation; that is,
p(S) is the zero operator when S is formally substituted in p(k).
Prove this in the case of the shape operator by showing that
SvSw  2HSvw + Kvw =
for any pair of tangent vectors to M .
The realvalued functions
I(v,w) = vw, II(v,w) = Svw,
and
III(v,w) = S vw = SvSw,
defined for all pairs of tangent vectors to an oriented surface, are
traditionally called the first, second, and third fundamental forms of M.
They are not differential forms; in fact, they are symmetric in v and w
rather than alternate. The shape operator does not appear explicitly in
the classical treatment of this subject; it is replaced by the second fun
damental form.
5. (Dupin curves). For a point p of an oriented region of M, let C be the
intersection of M near p with its tangent plane f p (M); specifically, C
consists of those points of M near p which lie in the plane through p
orthogonal to U(p). C may be approximated by substituting for M its
quadratic approximation M; thus Co is approximated by the curve
C : hx 2 + k*y 2 = 0, near (0, 0).
(a) Describe Co in each of the three cases K(p) > 0, K(p) < 0, and
K(p) = (not planar).
(b) Repeat (a) with Co replaced by C e and C_e, where the tangent
plane has been replaced by the two parallel planes at distance ±e
from it.
(c) This scheme fails for planar points since the quadratic approxima
tion becomes M : z = 0. For the monkey saddle, sketch C , Ce, and
C_e.
Sec 3]
GAUSSIAN CURVATURE
209
FIG. 5.23
6. If v, w, a, and b are vectors in E , prove the Lagrange identity
v»a v»b
(v X w).(a Xb) =
w»a w»b
7. (Parallel surfaces). Let M be a surface oriented by U; for a fixed number
e (positive or negative) let F: M — ► E 3 be the mapping such that
F(p) = p + eC/(p).
(a) If v is tangent to M at p, show that v = F* (v) is v — eS(\). De
duce that
where
v X w = «/(p) v X w,
J = 1  2etf + e 2 K = (1  efci)(l  ek 2 ).
If the function «/ does not vanish on M (M is compact and  e  small),
this shows that F is a regular mapping, so the image
M = F(M)
is at least an immersed surface in E 3 (Ex. 10 of IV.8) . M is said to
be parallel to M at distance e (Fig. 5.23).
(b) Show that the canonical isomorphisms of E make U a unit normal
on M for which S(r) = S(\).
(c) Derive the following formulas for the Gaussian and mean curvatures
of M:
K(F) = K/J; S(F) = (H  eK)/J.
8. (Continuation)
(a) Check the results in (c) in the case of a sphere of radius r oriented
by the outward normal U. Describe the mapping F = Fe when e
is 0, —r, and — 2r.
210 SHAPE OPERATORS [Chap. V
(b) Starting from an orientable surface with constant positive Gaussian
curvature, construct a surface with constant mean curvature.
4 Computational Techniques
We have denned the shape operator S of a surface M in E and found
geometrical meaning for its main algebraic invariants: Gaussian curvature
K, mean curvature H, principal curvatures fci and k 2 , and (at each point)
principal vectors ei and e 2 . We shall now see how to express these invariants
in terms of patches in M .
If x: D — * M is a patch inlc E 3 , we have already used the three real
valued functions
E = X U »X U , F = X U *X V = x v *x u , G = x„«x„.
on D. Here E > and G > are the squares of the speeds of the u and
yparameter curves of x, and F measures the coordinate angle # between
x u and x„, since
F = x u »x v —  x u   x„  cos # = s/EG cos &.
(Fig. 5.24). E, F, and G are the "warping functions" of the patch x: They
measure the way x distorts the flat region D in E 2 in order to apply it to
the curved region x(D) in M. These functions completely determine the
dot product of tangent vectors at points of x(D), for if
v = Vix u + v 2 x v and w = Wix u + w 2 x u ,
then
vw = EviWi + F{viw 2 + v 2 Wi) + Gv 2 w 2 .
(In such equations we understand that x„, x„, E, F, and G are evaluated at
(u, v) where x(u,v) is the point of application of v and w.)
Now x M X x„ is a function on D whose value at each point (u, v) of D
is a vector orthogonal to both x u (u, v) and x v (u, v) — and hence normal to
M at x(u, v). Furthermore, by Lemma 1.8 of Chapter II,
 Xu X x„  2 = EG F\
FIG. 5.24
Sec. 4] COMPUTATIONAL TECHNIQUES 211
Since x is, by definition, regular, this realvalued function on D is never
zero. Thus we can construct the unit normal function
j j X u y\ X v
Xi4 P\ Xp
on D, which assigns to each (u, v) in D a unit normal vector at x(u, v). We
emphasize that in this context, U, like x u and x„, is not a vector field on
x(D), but merely a vector valued function on D. Nevertheless we may
regard the system x u , x v , U as a kind of defective frame field. At least U
has unit length and is orthogonal to both x u and x„, even though x„ and
x v are generally not orthonormal.
In this context, covariant derivatives are usually computed along the
parameter curves of x, where by the discussion in Section 1, they reduce to
partial differentiation with respect to u and v. As in the case of x u and x„,
these partial derivative are again denoted by subscripts u and v. If
x{u, v) = (xi(u, v), x 2 (u, v), Xz(u, v)),
then just as for x„ and x„ on page 134, we have
(d 2 Xi d X2 d Xz\
_ ( d 2 £i d 2 X2 d Xz \
Xvu ~ \dudv ' dudv ' dudv/x
(d 2 Xi d 2 X2 d 2 X 3 \
Evidently x uu and x vv give the accelerations of the u and ^parameter
curves. Since order of partial differentiation is immaterial, x uv = x„„, which
gives both the covariant derivative of x M in the x„ direction, and x„ in the
x u direction.
Now if 8 is the shape operator derived from U, we define three more
realvalued functions on D:
( = S(x u )»x u
m = S(x u )*X v = S(x v )*X u
n = S(X V )'X V .
Because x u , x„ gives a basis for the tangent space of M at each point of
x(D), it is clear that these functions uniquely determine the shape operator.
Since this basis is generally not orthonormal, f, m, and n do not lead to
simple expression for S(x u ) and *S(x„) in terms of x u and x„. In the formulas
preceding Corollary 3.5, however, they do provide simple expressions for
Gaussian and mean curvature.
212
SHAPE OPERATORS
[Chap. V
4.1 Corollary If x is a patch in M CZ E , then
K(x) =
L 
EG  F 2 '
H(x) =
Gl + En  2F„
2(EG  F*)
Proof. At a point p of x(D), the formulas on page 206 express K(p) and
H(p) in terms of tangent vectors V(p) and W(p) at p. If V(p) and W(p)
are replaced by the tangent vectors x u (u, v) and x v (u, v) at x(u, v), we
find the required formulas for K(x(u, v)) and H(x(u, v)). I
When the patch x is clear from context, we shall usually abbreviate the
composite functions K(x) and H(x) to merely K and H.
By a device like that used in Lemma 2.1 we can find a simple way to
compute I, m , and n — and thereby K and i7. For example, since U*x u = 0,
partial differentiation with respect to v — that is, ordinary differentiation
along vparameter curves — yields
= — (U'x u ) = U v *x u + U'x uv .
dv
(Recall that U v is the covariant derivative of the vector field v — > U(u , v)
on each yparameter curve, u = uo.) Since x v gives the velocity vectors
of such curves, Exercise 1 of Section 1 shows that U v = —S(x v ). Thus the
preceding equation becomes
S(x v )x u = U'x uv .
(Fig. 5.25). Three similar equations may be found by replacing u by v,
and v by u. In particular,
S(x u )'X v = U'X VU = U*x uv = S(x v )*x u .
FIG. 5.25
Sec. 4]
COMPUTATIONAL TECHNIQUES
213
FIG. 5.26
Again, since x„ and x„ give a basis for the tangent space at each point, this
is sufficient to prove that S is symmetric (Lemma 1.4).
4.2 Lemma If x is a patch inlc E 3 , then
i = S(x u )'X u = U'Xuu
m = S(x u )»X v = U*X UV
n = S(x v )*X v = U'X VV .
The first equation in each case is just definition, and u and v may be
reversed in the formulas for m .
4.3 Example Computation of Gaussian and mean curvature
(1) He/i'co/d (Exercise 7 of Section 2, Chapter IV). This surface H,
shown in Fig. 5.26, is covered by a single patch
x(u, v) — {u cos v, u sin v, bv), 6^0,
for which
x M = (cos v, sin v, 0)
x„ = ( — u sin v, u cos v, b)
E = 1
F =
G = b 2 + u\
Hence
x„ X x„ = (b sin v, —b cos v, u).
214 SHAPE OPERATORS [Chap. V
To find K alone it is not necessary to compute E, F, and G, but it is
wise to do so anyway, since the identity
1 x u Xx,= VEG  F 2
then gives a check on the length of x u X x„. (Its direction may also be
checked, since it must be orthogonal to both x u and x„.) If we denote
 x u X x„  by W, then W = \/b 2 + u 2 for the helicoid, so the unit normal
function is
TJ _ x M X x„ (6 sin v, — b cos v, u)
w v& 2 +
Next we find
x«« u
x u » = ( — sin v, cos v, 0)
x„„ = (— u cos v, —u sin v, 0).
Here x uu = is obvious, since the wparameter curves are straight lines.
The vparameter curves are helices, and this formula for the acceleration
x vv was found already in Chapter II. Now by Lemma 4.2,
„ (, x w X Xd / ^
t = x
w
\ x u X x„ )
b
W
W
\ x u X x„ )
n
Hence by Corollary 4.1 and the results above,
L  m 2 (b/W) 2 b 2 b 2
K =
EG  F 2 W 2 W i (b 2 + u 2 ) 2
= Gl+ E n  2F m =
2(EG  F 2 )
Thus the helicoid is a minimal surface with Gaussian curvature
1 ^ K < 0.
The minimum value K = — 1 occurs on the central axis (u = 0) of the
helicoid, and K — > as distance  u \ from the axis increases to infinity.
(2) The saddle surface M: z = xy (Example 1.3). This time we use
the Monge patch x(u, v) = (u, v, uv) and with the same format as above,
compute
Sec. 4]
x M = (1, 0, v)
x v = (0, 1, u)
COMPUTATIONAL TECHNIQUES
E = 1 + v 2
F = uv
215
U = (v, u,l)/W
x uv = (0,0, 1)
G = 1 + M 2
TF = Vl + W 2 + V 2
£=
m = 1/W
„ = 0.
Hence
# =
1
(1 + u 2 + ?; 2 ) 2 '
# =
— uv
(1 + M 2 + V 2 )
3/2
Strictly speaking, these functions are K{x) and #(x) denned on the
domain E 2 of x. But it is easy to express K and H directly as functions on M
by using the cylindrical coordinate functions r = \/x 2 + y 2 and z. Note
from Fig. 5.27 that
r{x{u,v)) = \/u 2 + v 2
and
hence on M :
z(x(u, v)) = uv;
K =
1
H =
— z
(1 + r 2 ) 2 ' (1 +r 2 ) 3 ' 2
Thus the Gaussian curvature of M depends only on distance to the z
P = x(u, v)
FIG. 5.27
216 SHAPE OPERATORS [Chap. V
axis, rising from K = — 1 (at the origin) toward zero as r goes to infinity,
while H varies more radically.
Like all simple (that is, onepatch) surfaces, the helicoid and saddle
surface are orientable, since computations as above provide a unit normal
on the whole surface. Thus the principal curvature functions ki and k% are
defined unambiguously on each surface. These can always be found from
K and H by Corollary 3.5. Since the helicoid is a minimal surface, we get
the simple result
fci, h, =
For the saddle surface,
ki, h> =
(6 2 + w 2 )
z db Vl + r 2 + z 2
(1 + r 2 ) 3 ' 2
Techniques for computing principal vectors are left to the exercises.
There is a different computational approach which depends on having
an explicit formula Z = ^ ZiUi for a non vanishing normal vector field Z
on M . The main case is a surface given in the form M : g = c, for there we
know from Chapter IV, Section 3, that the gradient
dXi
is such a vector field — thus we may use any convenient scalar multiple of
Vgf as Z. Let S be the shape operator derived from the unit normal
u = z/\\ z .
If V is a tangent vector field on M, then by Method 2 in Section 1, we find
VyZ = X) V[Zi]Ui.
Using a Leibnizian property of such derivatives,
V V C7 = V,
Z (VyZ)
+ F [m] ;
(Fig. 5.28). What is important here is that V[l/\\ Z \\]Z is a normal vector
field; we do not care which one it is, so we denote it merely by — N v . Thus
5(F) v vU = =l£p + Ny.
Note that if W is another tangent vector field on M, then N v X N w = 0,
while products such as N v X Y are tangent to M for any Euclidean vector
Sec. 4] COMPUTATIONAL TECHNIQUES 217
Tangent plana
V[l/\\ Z \\]Z
FIG. 5.28
field Y on M. Thus it is a routine matter to deduce the following lemma from
Lemma 3.4.
4.4 Lemma Let Z be a nonvanishing normal vector field on M . If V and
W are tangentvector fields such that V X W = Z, then
R = (ZVyZ X V W Z)
H = Z>
II Z  4
(v v z x w + V X V W Z)
211 Z II 3
To compute, say, the Gaussian curvature of a surface M: g = c using
patches, one must begin by explicitly finding enough of them to cover all of
M; a complete computation of K may thus be tedious, even when g is a
rather simple function. The following example shows to advantage the
approach just described.
4.5 Example Curvature of the ellipsoid
2 2 2
M: 9 =  2 +  +  2 = 1.
a 2 o 2 c 2
We write g = J2 Xi/ai, and use the (nonvanishing) normal vector field
Z = ^g=j:^U i .
at
Now if V = J] «iC7, is a tangent vector field on M,
«i a» 2
since
218
SHAPE OPERATORS
[Chap. V
V[xi] = dxt(V) = vt.
Similar results for another tangent vector field W then yield
z*v v z x v w z =
Xi
X 2
«2 2
x 3
a 3 2
Vi
«i 2
V2
a 2 2
Vz
a 3 2
a 2 2
w 3
a 3 2
1
ai 2 a 2 2 a 3 2
XV X w.
where X is the special vector field ^ XiUi which was used in Example 3.9
in Chapter IV.
It is always possible to choose V and W so that V X W = Z.
But then
2
X*V X W = ZZ = £— = !•
a* 2
Thus by Lemma 4.4 we have found
1
K =
afatW II Z  4
where
"(^7
For an?/ oriented surface in E , its support function h assigns to each
point p the orthogonal distance h(p) = p»£/(p) from the origin to the
tangent plane T P (M), as shown in Fig. 5.29 for the ellipsoid. Using the
vector field X (whose value at p is the tangent vector p p ), we find for the
ellipsoid that
<7 1
h = X'U = X,
z
FIG. 5.29
Sec. 4] COMPUTATIONAL TECHNIQUES 219
Thus a more intuitive expression for the Gaussian curvature of the ellipsoid
is
K= h *
a 2 b 2 c 2
Note that when a = b = c = r, the ellipsoid is a sphere and this formula
becomes K = 1/r 2 .
The computational results in this section, though stated for surfaces,
still apply to immersed surfaces (Exercise 10 of Chapter IV, Section 8).
In particular, the formulas in Corollary 4.1 make sense for an arbitrary
regular mapping x: D — > E 3 . The theoretical justification of this added
generality is outlined in Chapter VII, Section 7.
EXERCISES
1. Show that the sphere of radius r has K = 1/r 2 by applying the methods
of this section to the geographical patch
x(u, v) = (r cos v cos u, r cos v sin u, r sin v).
2. For a Monge patch, x(u,v) = (u, v,f(u,v)), show that
E = 1 + fj t = f uu /W
F = fufv m = fuv/W
G = 1 + jv n = U/w
where
W = (1 +/„*■+ /."J 1 *.
Find formulas for K and H.
3. (Continuation). Deduce that the image of x is flat if and only if
Juujvv Juv == "J
minimal if and only if
(1 + fu)f vv + (1 + /„ 2 ) /„„  2f u f v f uv = 0.
4. Show that the image of the patch
x(w, v) = (u, v, log cos v — log cos u)
is a minimal surface with Gaussian curvature
220 SHAPE OPERATORS [Chap. V
„ _ —sec u secv
where
W 2 = 1 + tan 2 w + tanV
5. Express the curvature K of the monkey saddle M: z = x 3 — 3xy 2
(Fig. 4.47) in terms of r = s/x 2 + y 2 . Is this surface minimal?
6. Find the Gaussian curvature of the elliptic and hyperbolic paraboloid,
M:z = 2 + 8 C> < e = ±D
7. Show that the curve segment
a(t) = x(d(0, <h(t)), a ^ t g b
has length
L(a) = f (Eai 2 + 2Fai a 2 ' + Ga^) m dt,
•'a
where 2?, i^, and (? are evaluated on 0,1,02.
8. Prove that the coordinate angle # of a patch x: D — > M, < # < ir,
is a differ entiahle function on D. (Hint: Use the Schwarz inequality in
III)
9. (a) A patch x in M is orthogonal provided F = (so x u and x„ are
orthogonal at each point ) . Show that in this case
&\Xu) == "^ X M f ^ X v
0\X V ) = ^ X u p — X„ .
(b) A patch x is principal provided F = m = 0. Prove that x„ and x„
are principal vectors at each point, with corresponding principal
curvatures l/E and n/G.
10. Prove that a tangent vector
V = ViX u + v 2 x v
is a principal vector if and only if
11 2
= 0.
2
V2
— Viv 2
2
jE7
F
G
f
m
n
(Hint: v is principal if and only if the normal vector aS(v) X v is zero.)
Sec. 4] COMPUTATIONAL TECHNIQUES 221
11. Show that on the saddle surface M (4.3) the two vector fields
(Vi + u\ ± Vi + v 2 , i>Vi + u 2 ± wVi + y 2 )
are principal at each point. Check that they are orthogonal and tangent
toM.
12. (Enneper's minimal surface). This is the immersed surface given by
t \ ( U . 2 *> 3 . 2 2 2\
x(w, v) = iu — — + UV , V — — + uv,u — V \.
Prove that this immersed surface is minimal and that x is not oneto
one. {Hint: For H = it suffices to prove
E = G, F = 0, and x uu + x„„ = 0.)
13. (Patch criterion for umbilics). Show that the point x(u, v) is umbilic
if and only if there is a number k such that t = kE, m = kF, and
«. = kG at (u, v) (A; is then the principal curvature ki = k%).
14. If v = vix u + V2X„ is tangent to M at x(w, v), the normal curvature in
the direction determined by v is
jf( \ _ ^ + 2mViV 2 + ««2 2
~ j&yi 2 + 2FviV 2 + Gy 2 2
where the various functions are evaluated at (u,v).
15. Find the umbilic points (if any) on the following surfaces:
(a) Saddle (Example 4.3).
(b) Monkey saddle (Exercise 5).
(c) Elliptic paraboloid (Exercise 6).
16. (Tubes). If /? is a unitspeed curve in E 3 with k > 0, let
x(u,v) = j3(u) + e(cos v N(u) + sin vB(u)).
Thus the t;parameter curves are circles of (constant) radius e in planes
orthogonal to /?. Show that
(a) x is regular if e is small enough; so x is an immersed surface called
the tube' of radius e around /3.
(b) U = cos v N(u) + sin v B(u) is a unit normal function on the
tube.
/ \ V —k(u) cos V
\c) J\. =
e(l — k(u) e cos v)
(Hint: Use S(x u ) X S(x v ) = Kx u X x„.)
17. Show that the elliptic hyperboloids of one and two sheets (Ex. 10 of
IV.2) have Gaussian curvature K = —h*/a 2 b 2 c 2 and K = h /a 2 b 2 c 2 , re
222
SHAPE OPERATORS
[Chap. V
spectively, and that both support functions h are given by the same
formula as for the ellipsoid (4.5).
18. If h is the support function of an oriented surface M (Z E 3 , show that
(a) A point p of M is a critical point of h if and only if pS(x) —
for all tangent vectors v to M at p. (Hint: Write h as X* U, where
X = J^XiUi.)
(b) When K(p) ^ 0, p is a critical point of h if and only if p (con
sidered as a vector) is orthogonal to M at p.
19. Use the preceding exercises to find the Gaussian curvature intervals
of the ellipsoid and the elliptic hyperboloids of one and two sheets.
(Ex. 10 of IV. 2) Assume a ^ b ^ c.
20. Compute K and H for the saddle surface (Example 4.3) by the method
given at the end of this section. (Hint: Take V and W tangent to the
two sets of rulings of M. )
21. Scherk's minimal surface, M: e cos x = cos y. Let (R be the region
in the xy plane on which cos x cos y > 0; (R is a checkerboard pattern
of open squares, with vertices ((x/2) + irm, Or/ 2 ) + nir ) Show that
(a) M is a surface.
(b) For each point (u,v) in (R there is exactly one point (u, v, w) in M .
The only other points of M are entire vertical lines over each of
the vertices of (R (Fig. 5.30).
(c) M is a minimal surface with K = e z /(e z mxx + l) 2 . (Hint:
FIG. 5.30
Sec. 5] SPECIAL CURVES IN A SURFACE 223
V = cos xUi + sin z£/ 3 is a tangent vector field.) Further proper
ties of this surface are given in Ex. 1 1 of VI.8.
22. Let Z be a neverzero normal vector field on M. Show that a tangent
vector v to M at p is principal if and only if
vZ(p) XV,Z =
(see the hint for Exercise 10).
The equation above together with the tangency condition
Z(p)v =
may be solved for the principal directions. Thus umbilics may be
located using these equations: p is umbilic if and only if every tangent
vector v at p is principal.
23. Consider the ellipsoid M : ]£ x?/a? = 1. Show that
(a) A tangent vector v at the point p is principal if and only if
= piV2V S ((i2 2 — (h) + P2ViV 3 (a 3 2 — a?) + PzV\Vi(ai — a*).
(b) Assuming a x > a 2 > a 3 , show that there are exactly four umbilics
on M, with coordinates
2\ 1/2 / 2 2\ 1/2
# tti — a
pi = ±ai
(2 2\ 1/2 / 2 „ 2\ 1
ai — a 2 \ n . /a2 — a 3 \
5 Special Curves in a Surface
We shall briefly consider three geometrically significant types of curves in a
surface M C E 3 . Neither this section nor the next is really essential for the
theory; their purpose is to illustrate some of the ideas already introduced,
and supply examples for later work.
5.1 Definition A regular curve a in M C E 3 is a principal curve (or line
of curvature) provided that the velocity a of a always points in a principal
direction.
Thus principal curves always travel in directions in which the bending
of M in E 3 takes its extreme values. Neglecting changes of parametrization,
there are exactly two principal curves through each nonumbilic point of
M— and these necessarily cut orthogonally across each other. (At an umbilic
point p, every direction is principal and near p the pattern of the principal
curves can become quite complicated.)
5.2 Lemma Let a be a regular curve in M CI E 3 , and let U be a unit
normal vector field restricted to a. Then
224
SHAPE OPERATORS
[Chap. V
(1 ) The curve a is principal if and only if U and a are collinear at each
point.
(2) If a is a principal curve, then the principal curvature of M in the
direction of a is a" • U/a*a.
Proof. (1) Exercise 1 from Section 1 shows that S(a) = — U'. Thus
U and a are collinear if and only if S(a ) and a are collinear. But by
Theorem 2.5, this amounts to saying that a always points in a principal
direction, or, equivalently, that a is a principal curve.
(2) Since a is a principal curve, the vector field a / a  consists entirely
of (unit) principal vectors belong to, say, the principal curvature ki.
Thus
h = *(«7 «' ) =S(a'/\\a'\\)a'/\\a'\\.
= S(a')a = a"U
a! *a' a' *a'
where the last equality uses Lemma 2.1. 
In this lemma, (1 ) is a simple criterion for a curve to be principal, while
(2 ) gives the principal curvature along a curve known to be principal.
5.3 Lemma Let a be a curve cut from a surface M C E by a plane P.
If the angle between M and P is constant along a, then a is a principal
curve of M.
Proof. Let U and V be unit normal vector fields to M and P (respectively)
along the curve a, as shown in Fig. 5.31. Since P is a plane, V is parallel,
that is, V = 0. Now the constant angle assumption means that U»V is
constant; thus
= (U'V)' = U''V.
Since U is a unit vector, U' is orthogonal to U as well as V. The same is of
FIG. 5.31
Sec. 5] SPECIAL CURVES IN A SURFACE 225
course true of a, since a lies in both M and P. If
U and F are linearly independent (as in Fig.
5.31) we conclude that U' and a are collinear;
hence by Lemma 5.2, a is principal.
However, linear independence fails only when
U = ±F. But then if = 0, so a is (trivially)
principal in this case as well. 
(It is scarcely any harder to prove the gener
alization given in Exercise 5.) Using this result it FIG  5  32
is easy to see that the meridians and parallels of a
surface of revolution M are its principal curves. Indeed, each meridian ju is
sliced from M by a plane through the axis of revolution and hence orthog
onal to M along n, while each parallel ir is sliced from M by a plane orthog
onal to the axis, and by rotational symmetry such a plane makes a con
stant angle with M along ir.
Directions tangent to M C E 3 in which the normal curvature is zero are
called asymptotic directions. Thus a tangent vector v is asymptotic provided
fc(v) = /S(v)»v = 0, so in an asymptotic direction M is (instantaneously, at
least) not bending away from its tangent plane.
Using Corollary 2.6 we can get a complete analysis of asymptotic direc
tions in terms of Gaussian curvature.
5.4 Lemma Let p be a point of M C E .
(1) If K(p) > 0, there are no asymptotic directions at p.
(2) If K(p) < 0, then there are exactly two asymptotic directions at p
which are bisected by the principal directions (Fig. 5.32) at angle & such
that
tan 2 * = :=ME } .
fe(p)
(3) If K(p) = 0, then every direction is asymptotic if p is a planar
point; otherwise there is exactly one asymptotic direction which is also
principal.
Proof. These cases all derive from Euler's formula
A;(u) = h(p) cos 2 # + fc 2 (p) sinV
in Corollary 2.6.
(1) Since fci(p) and A; 2 (p) have the same sign, fc(u) is never zero.
(2) Here fci(p) and /^(p) have opposite signs and we can solve the
equation = fci(p) cosV + A 2 (p) sin 2 ?? to obtain the two asymptotic
directions.
(3 ) If p is planar, then
226
[Chap. V
FIG. 5.33
h(p) = k»(p) = 0;
hence Zs(u) is identically zero. If just
then
fc 2 (p) = 0,
/c(u) = fci(p) cos 2 #
is zero only when cos & = 0, that is, in the principal direction u = e 2 . 
We can get an approximate idea of the asymptotic directions at a point
p of a given surface M by picturing the intersection of the tangent plane
f p (M) with M near p. When K (p) is negative, this intersection will consist
of two curves through p whose tangent lines (at p ) are asymptotic direc
tions (Exercise 5 of Section 3 ) .
Figure 5.33 shows the two asymptotic directions A and A at a point p
on the inner equator of a torus. (The two intersection curves merge into a
single figure8.)
5.5 Definition A regular curve a in M <Z E 3 is an asymptotic curve pro
vided its velocity a always points in an asymptotic direction.
Thus a is asymptotic if and only if
k(a) = S(a)a = 0.
Since S(a) = — U', this gives a criterion, U' »a = 0, for a to be asymp
totic. Asymptotic curves are more sensitive to Gaussian curvature than are
principal curves: Lemma 5.3 shows that there are none in regions where K
is positive, but two cross (at an angle depending on K) at each point of a
region where K is negative.
The simplest criterion for a curve in M to be asymptotic is that its ac
celeration a" always be tangent to M. In fact, differentiation of U*a =0
gives
Sec. 5] SPECIAL CURVES IN A SURFACE 227
U'a + Ua" = 0,
so U'»a — (a asymptotic) if and only if U*a" = 0.
The analysis of asymptotic directions in Lemma 5.4 has consequences
for both flat and minimal surface. First, a surface M in E is minimal if
and only if there exist two orthogonal asymptotic directions at each of its
points. In fact, H (p ) = is equivalent to fci (p ) = — k 2 (p ) , and an examina
tion of the possibilities in Lemma 5.4 shows that /bi(p) = — A^(p) if and
only if either (a) p is planar (so the criterion holds trivially) or (b)
K(p) < withtf = ±x/4,
which means that the two asymptotic directions are orthogonal.
Thus a surface is minimal if and only if through each point there are two
asymptotic curves which cross orthogonally. This observation gives geomet
ric meaning to the calculations in Example 4.3 which show that the helicoid
is a minimal surface. In fact, the u and vparameter curves of the patch x
are orthogonal since F = 0, and their accelerations are tangent to the surface
since £ = U*x uu = and n = U*x vv = 0.
Roughly speaking, a ruled surface M is one which is swept out by a straight
line moving through E — the various positions of the line are called the
rulings of M. Thus M has a parametrization in the ruled form
x(w, v) = j8(w) + v8(u), or /3(v) + u8(v)
where /3 and 5 are curves in E with 5 never (see Exercise 49 of Section 2
of Chapter IV). For example, the helicoid is a ruled surface, since the
patch in Example 4.3 may be written as
x(w, v) = (0, 0, bv) + w(cos v, sin v, 0).
This shows how the helicoid is swept out by a line rotating as it rises along
the z axis. The saddle surface M : z = xy is doubly ruled, since
, x / x f(u, 0, 0) + v(0, l,u)
It is no accident that both these surfaces have K negative, for:
5.6 Lemma A ruled surface M has Gaussian curvature K ^ 0. Further
more K = if and only if the unit normal U is parallel along each ruling of
M (so all points p on a ruling have the same tangent plane f p (M).)
Proof. A straight line t —> p + tq in any surface is certainly asymptotic,
since its acceleration is zero, and thus trivially tangent to M . By definition,
a ruled surface contains a straight line through each of its points. Hence
there is an asymptotic direction at each point, so by Lemma 5.4, K fS 0.
228 SHAPE OPERATORS [Chap. V
Now let a (t) = p + tq be an arbitrary ruling in M. If U is parallel along
a, then S (a ) = U = 0. Thus a is a principal curve with principal curva
ture k(a) = 0, and so K = k^ = 0.
Conversely, if K = 0, we deduce from the case (3) in Lemma 5.4 that
asymptotic directions (and curves) in M are also principal. Thus each rul
ing a is principal (S(a) = k(a)a) as well as asymptotic (k(a) = 0):
hence U' = S(a) = 0. 
We come now to the last and most important of the three types of curves
under discussion.
5.7 Definition A curve a in M <Z E 3 is a geodesic of M provided its ac
celeration a" is always normal to M.
Since a" is normal to M , the inhabitants of M perceive no acceleration at
all — for them the geodesic a is a "straight line." A full study of geodesies
is given in Chapter VII, where in particular we examine their character as
shortest routes of travel. Geodesies are far more plentiful on a surface M
than are principal or asymptotic curves; indeed by Theorem 4.2 of Chapter
VII there is a geodesic through every point of M in every direction.
Since its acceleration a" is, in particular, orthogonal to its velocity a, a
geodesic a has constant speed, for differentiation of
 a  2 = a *a yields 2a •«" = 0.
A straight line a(t) = p + U\ in M is always a geodesic of M, since its
acceleration a" = is trivially normal to M. Unlike principal and asymp
totic curves, geodesies cannot be defined in terms of the shape operator;
however, a (unit speed) geodesic a with positive curvature bears an in
teresting relation to S, which uses the Frenet apparatus of a. Since the
principal normal N = a" '/k of a is normal to the surface M, we have
N' = S(a) = S(T).
Thus by a Frenet equation, S(T) = kT  tB.
These remarks are sufficient to derive the geodesies of three rather special
surfaces.
5.8 Example Geodesies on some surfaces in E .
(1) Planes. If a is a geodesic in a plane P orthogonal to u, a «u = 0,
hence a" *u = 0. But a" is by definition normal to P; hence a." = 0. Thus
a is a straight line, and since every such line is a geodesic, we conclude
that the geodesies of P are all straight lines in P.
(2) Spheres. If a is a (unit speed) geodesic in a sphere S of radius r,
then, by a remark above, S(T) = kT — tB. (We saw in Chapter II, Sec
tion 3, that any curve in S has positive curvature, so the Frenet apparatus
Sec. 5]
SPECIAL CURVES IN A SURFACE
229
FIG. 5.34
is available.) But Example 1.3 shows that S(T) = ±(\/r)T, depending
on which of the two unit normals is used. These two equations for S(T)
imply that k = \/r and t = 0. Hence by Lemma 3.6 of Chapter II, a lies
on a circle C of radius r. This maximum radius r forces C to be a great circle
of 2, that is, one sliced from 2 by a plane through its center. Conversely,
any constantspeed curve running along a great circle has its acceleration
a" point toward the center of that circle, which is also the center of 2, so
a." is normal to 2. We conclude that the geodesies of 2 are the constantspeed
parametrizations of its great circles (Fig. 5.34).
(3) Cylinders. The geodesies of, say, the circular cylinder M:x 2 \ y 2 = r 2
are all curves of the form
a(t) = (r cos (at + b), r sin (at + b), ct + d).
In fact, any curve in M may be written
a(t) = (rcos t?(0, r sin d(t), h(t)),
and a vector normal to M has zcoordinate zero. Thus if a is a geodesic,
then h" = 0, so h(t) = ct + d. The speed (rV 2 + fi*) 1 ' 2 of a is constant,
so & is constant; hence &(t) = at + 6.
When the constants a and c are both nonzero, a is a helix on M. The
extreme case a = gives the rulings of M, and c = gives the cross
sectional circles.
The essential properties of the three types of curves we have considered
may be summarized as follows :
Principal curves k (a ) = ki or k 2 S (a ) collinear a
Asymptotic curves k(a) = S (a) orthogonal a a" tangent M
Geodesies a" normal M
EXERCISES
1. Prove that a curve a in M is a straight line of E 3 if and only if a is
both geodesic and asymptotic.
230
SHAPE OPERATORS
[Chap. V
2. To which of the three types — principal, asymp
totic, geodesic — do the following curves belong?
(a) The top circle a on the torus (Fig. 5.35).
(b) The outer equator /3 of the torus.
(c) The x axis on M : z = xy.
(Assume a constantspeed parametrization. )
3. On a surface of revolution, show that all meridians are geodesies, but
that the parallel through a point a(t) of the profile curve is a geodesic
if and only if a (t) is parallel to the axis of revolution.
4. Let a be an asymptotic curve in M <Z E 3 with curvature k > 0.
(a) Prove that the binormal B of a is normal to the surface along a,
and deduce that S(T) = tN.
(b) Show that along a the surface has Gaussian curvature K = — r 2 .
(c) Use (b) to find the Gaussian curvature of the helicoid (Example
4.3).
5. Suppose that a curve a lies in two surfaces M and M which make a
constant angle along a ([/• £7 constant). Show that a is principal in M
if and only if principal in M.
6. If x is a patch in M, prove that a curve a(t) = x(ai(t), cte(0) is
(a) Principal if and only if
E
/2
— ai ai a\
F G
= 0,
(b) Asymptotic if and only if fa/ 2 + 2mai'a 2 / + n.a^ = 0.
Let a be a unitspeed curve inlc E 3 . Instead of the Frenet frame
field on a, consider the frame field T, V, , U — where T is the unit tangent
of a, U is the surface normal restricted to a, and V = U X T (Fig.
5.36).
(a) Show that
T' = gV + kU U
V' = gT + tU
U' = kT  tV.
where k = S(T)*T is the nor
mal curvature k(T) of M in the
T direction, and t = S(T)'V.
FIG. 5.36
Sec. 5] SPECIAL CURVES IN A SURFACE 231
The new function g is called the geodesic curvature of a.
(b) Deduce that a is
geodesic <=» g =
asymptotic <=> fc =
principal <=> tf = 0.
8. If a is a (unit speed) curve in M, show that
(a) a is both principal and geodesic if and only if it lies in a plane
everywhere orthogonal to M along a.
(b) a is both principal and asymptotic if and only if it lies in a plane
everywhere tangent to M along a.
9. On the monkey saddle M (Ex. 5 of IV.4) find three asymptotic curves
and three principal curves passing through the origin 0. (This is possi
ble because only because is a planar umbilic point.)
10. Show that the ruled surface x(u, v) = /3(w) + v8(u) has Gaussian
curvature
K== m 2 (js'a x b'y
EG  F 2 W 4
where
W =/?' X 8 + v8' X«.
11. (Flat ruled surfaces).
(a) Show that cones and cylinders are flat (see Exs. 5 and 6 of IV.2)
(b) If is a unitspeed curve in E 3 with k > 0, the ruled surface
x(u,v) = 0(w) + vT(u), v > 0,
is called the tangent surface of /? (Fig. 5.37). Prove that x is regu
lar, and that the tangent surface is flat.
FIG. 5.37
232 SHAPE OPERATORS [Chap. V
12. Let a be a regular curve in M C E 3 , and let U be the unit normal of
M along a. Show that a is a principal curve of Af if and only if the ruled
surface x(u, v) = a(u) + vU(u) is flat.
13. A closed geodesic of M is a geodesic which is a periodic function
a : R — * M . Find all closed geodesies in a sphere, a plane, and a circular
cylinder.
14. A ruled surface is noncylindrical if its rulings are always changing
directions; thus for any director curve, 8 X 8' ^ 0. Show that
(a) a noncylindrical ruled surface has a parametrization
x(u, v) = a(u) + v8(u)
for which  5  = 1 and a •$' = 0.
(b) for this parametrization,
v p\u) <x'b X 8'
& = / o/ x — ; — ^r« where p = ; — ; — .
(p\u) + v 2 ) 2 F 8 f »8'
The curve a is called the striction curve, and the function p is the
distribution parameter.
(Hint: For (a), if  5  = 1, find /so that a = a + f8. For (b), show
a X 5 = pS'.)
15. Describe the qualitative behavior of Gaussian curvature K on an
arbitrary ruling of a (noncylindrical) ruled surface. Show that the
route of the striction curve is independent of the choice of parametriza
tion, and that the distribution parameter is essentially a function on
the set of rulings.
16. Show that the striction curve of the helicoid is its central axis, and that
its distribution function is constant.
1 7. Find the striction curve and distribution parameter for
(a) Both sets of ruling of the saddle surface (Example 4.3)
(b ) Both sets of rulings of the hyperboloid of revolution
2,2 2
M: — = 1
a 2 b 2
(Fig. 5.38) (Find a ruled parametrizetion by modifying
Ex. 9 of IV.2.)
18. If x(u,v) = a(u) + v8(u) parametrizes a noncylindri
cal ruled surface, let L(u) be the ruling through a(u).
Show that
(a) If t? £ is the (smallest positive) angle from L(u) to
L(u + e), and
FIG. 5.38
Sec 5]
SPECIAL CURVES IN A SURFACE
233
d e is the orthogonal distance from L(u) to L{u + e), then lim (d E /
£»0
# e ) = p(w). Thus the distribution parameter is the reciprocal of
the rate of turning of L — its sign describes the direction of the
turning.)
(b) There is a unique point p E of L(u) which is nearest to L(u + e),
and lim p e = a(u). (This gives another characterization of the
£»0
striction curve.)
(c) The distance from a(u) to a(u + e) need not be a good approxi
mation of the distance d e from L(u) to L(u + e). Give an example.
19. Let x(u,v) = a + v8,  5  = 1, parametrize & flat ruled surface M.
Show that
(a) If a' is always zero, then M is a cone.f
(b) If h' is always zero, then M is a cylinder.
(c) If both a and 5' are never zero, then M is the tangent surface of
its striction curve. (Exercise lib.)
Of course these cases are far from exhausting all the possibilities,
but in a sense they show that an arbitrary flat ruled surface ("de
velopable surface") is a mixture of the three types in Exercise 11. If
such a surface is closed in E 3 (Ex. 10 of IV.8), it must be a cylinder,
since the closure condition implies that the rulings are entire straight
lines.
20. A right conoid is a ruled surface whose rulings all pass through a fixed
axis (Fig. 5.39). Taking this axis as the z axis of E 3 , we get the para
metrization
x(u,v) = (u cos $(v), u sin &(v), h(v))
(a) Find the Gaussian and mean curvature.
(b) Show that the surface is noncylindrical if
i?' is never zero; find its striction curve
and parameter of distribution.
21. Sketch the conoid M parametrized by
x(u, v) = (u cos v, u sin v, cos 2 v),
and find its Gaussian and mean curvature.
Express M in the form z = f(x,y) (origin
omitted).
22. Prove that a surface which is both ruled and fig. 5.39
minimal is part of either a plane or a helicoid.
t In each case, we assume there are no restriction on v except those necessary
to ensure that x is regular.
234 SHAPE OPERATORS [Chap. V
{Hint: Flat regions in M are planar; thus arguing as in Theorem 6.2
we may suppose K < 0. Use the parametrization in Exercise 14 with
the additional feature that 8 is a unitspeed curve. Then H = leads
to three equations. Deduce that 8 is a unit circle; we may assume
8(u) = (cos u, sin u, 0).)
6 Surfaces of Revolution
The geometry of a surface of revolution is rather simple, yet these surfaces
exhibit a wide variety of geometric behavior; thus they offer a good field
for experiment.
We shall apply the methods of Section 4 to study an arbitrary surface of
revolution M, parametrized as in Example 2.5 of Chapter IV by
x(u,v) = (g(u), h(u) cos v, h(u) sin v).
Recall that h(u) is the radius of the parallel of M at distance g(u) along
the axis of revolution, as shown in Fig. 4.17. This geometric significance of
g and h means that our results do not depend on the particular position of
M relative to the coordinate axes of E 3 .
Because g and h are functions of u alone, we write
x u = (</', h' cos v, h' sin v) E = g' 2 + h' 2
F =
x„ = (0, —h sin v, h cos v) G = h 2 .
Here E is the square of the speed of the profile curve — and of all meridians
(wparameter curves) — while G is the speed squared of the parallels (v
parameter curves). Next
x M X x„ = (hh , — hg cos v, —hg sin v)
 x u X x v  = VEG^F 2 = hVg' 2 + h' 2
U = (h', —g cos v, —g sin v)/vV 2 + h' 2 .
Taking second derivatives, we obtain
x uu = (g", h? cos v, h" sin v)
x™ = (0, — h sin v, h cos v)
Xw = (0, —h cos v, —h sin v)
I = igh" + g"h')/Vg' 2 + h' 2
m =
n = gh/Vg' 2 + h' 2
Sec. 6]
SURFACES OF REVOLUTION
235
Since F — m = 0, it is easy to check (Exercise 9 of Section 4) that for
the shape operator S derived from U,
0\X.u) J, Xi
ld\X v J — ™ \ v .
(Thus we have an analytical proof that the meridians and parallels of M
are its principal curves.) Hence if the corresponding principal curvature
functions are denoted by k„. and K (instead of k\ and fc 2 ), we have
g' h'
fCu — ^ —
g" h"
E (g' 2 + h' 2 ) 3 ' 2 '
Thus the Gaussian curvature of M is
/C?r — "7S —
9
K  Mv  T(gf2 + h , 2)2
G h(g' 2 + h' 2 ) 1 ! 2
g' h'
a" h"
(1)
(2)
Note that this formula defines K as a real valued function on the domain
/ of the profile curve
oc(u) = (g(u), h(u),0).
By the conventions in Section 4, K(u) is the Gaussian curvature K(x(u,v))
of every point on the parallel through a(u). Similarly, with the other functions
above — because of the rotational symmetry of M about the axis of revolu
tion, its geometry is "constant on parallels" and completely determined by
the profile curve.
In the special case when the profile curve passes at most once over each
point of the axis, we can usually arrange for the function g to be simply
g(u) = u (Exercise 13 of Chapter IV, Section 2). Thus the formulas
above reduce to
rCii —
k r =
K =
h"
(1 + h' 2 ) 3 ' 2
1
h(i + h' 2 y 2
v
(3)
h(l + h' 2 ) 2 '
6.1 Example Surfaces of Revolution
(1) Torus of revolution T. The parametrization x in Example 2.6 of
Chapter IV has
236 SHAPE OPERATORS [Chap. V
g(u) = r sin u, h(u) = R + r cos u.
Although the axis of revolution is now the z axis, formulas (1) and (2)
above remain valid and we compute
E = r 2 F = G = (R + r cos uf
£ = r m = n = (R + r cos u) cos u
*. = ! *,= cosw
r i? + r cos w
cos u
K =
r(R + r cos w) '
We now have an analytical proof that K is positive on the outer half
of the torus and negative on the inner half. In fact K has its maximum
value l/r(R + r) on the outer equator (u = 0) and its minimum value
l/r(R  r)
on the inner equator (u = T ), and K is zero on the top and bottom circles
(u = ±tt/2).
(2) Catenoid. The curve y = c cosh (x/c) is a catenary; its shape is that
of a chain hanging under the influence of gravity. The surface obtained by
revolving this curve about the x axis is called a catenoid (Fig. 5.40). From
the formulas in (3 ) we get
/C u = K T =
c cosh 2 (u/c)
Hence
K  1
c 2 cosh 4 (w/ c )
Since its mean curvature H is zero, the catenoid is a minimal surface.
Its Gaussian curvature interval is — 1/c 2 5j K < 0, with minimum value
& = — 1/c on the circle w = 0.
The following result shows that catenoids are the only complete] non
planar surfaces of revolution which are minimal. (A plane is trivially a
minimal surface, since k\ = fc 2 = 0. )
6.2 Theorem If a surface of revolution M is a minimal surface, then M
is contained in either a plane or a catenoid.
t We use this word with its dictionary meaning; a mathematical definition is
given in Section 4 of Chapter VII.
Sec. 6]
SURFACES OF REVOLUTION
237
FIG. 5.40
Proof. We use the parametrization
x(u, v) — (g(u), h(u) cos v, h(u) sin v)
of M, with u in some interval /, and v arbitrary.
Case 7. g is identically zero. Then g is constant, so M is part of a plane
orthogonal to the axis of revolution.
Cose 2. g is never zero. Then by an earlier exercise, M has a parametriza
tion of the form
y(u, v) = (u,f(u) cos v,f(u) sin v).
The formulas for k x and k 2 in (3) on page 235 then show that the minimal
ity condition ki + k 2 = is equivalent to
//" = 1+/' 2 .
Because u does not appear explicitly in this equation, there is a standard
method for solving it, which may be found in any elementary text on
differential equations. We merely record that the solution is
f(u) = a cosh
M
where a and b are arbitrary constants. Thus M is part of a catenoid.
Cose 3. g is zero at some points, nonzero at others. This cannot occur.
For definiteness suppose that g ' (u ) = 0, but g (u) > f or u < u . By
case 2, the profile curve
«(w) = (g(u), h(u),0)
238 SHAPE OPERATORS [Chap. V
is a catenary for u < u . But the shape of this curve makes it clear that its
slope h /g cannot suddenly become infinite at u = uq. 
Helicoids and catenoids are the "elementary" minimal surfaces. Two
others are given in Exercises 12 and 21 of Section 4. Soapfilm models of
an immense variety of minimal surfaces may easily be constructed by the
methods given in Courant and Robbins [4], where the term minimal is
explained.
The expression vV 2 + ^' 2 which appears so frequently in the general
formulas at the beginning of this section is, of course, just the speed of the
profile curve
«(«) = (9(u), h(u),0).
Thus we can radically simplify matters by replacing a by a unitspeed
reparametrization. The resulting surface of revolution M is unchanged:
We have merely given it a new parametrization, said to be canonical.
6.3 Lemma If x is a canonical parametrization of a surface of revolution
M (g 2 + ti 2 = 1), then
E = 1, F = 0, G = h 2
and
h"
K =
h
Proof. Since g' 2 + h! 2 = 1, these expressions for E, F, and G follow im
mediately from those on page 234, and K in (2) becomes
* = ^f
h'
_ g hr + g g"h
h
Differentiation of g 2 = 1 — h' 2 yields g g" = —h h" ; hence
„ (1  h ,2 )h"  h' 2 h" hT 
K = r = 7 • I
h h
The effect of canonical parametrization is to shift the emphasis somewhat
from measurements in the space outside M (for example, along the axis
of revolution) to measurements within M itself. This idea will be developed
more fully in Chapters VI and VII.
6.4 Example Canonical parametrization of the catenoid (c = 1).
An arclength function for the catenary
a(u) = (u, cosh u) is s(u) = sinh u;
Sec. 6] SURFACES OF REVOLUTION 239
}*(•)
9(s)
FIG. 5.41
hence a unitspeed reparametrization is
0( 8 ) = (g(s),h(s)) = (sinh _1 s, y/l + s 2 )
as indicated in Fig. 5.41. The resulting canonical parametrization of the
catenoid is then
x(s, t>) = (sinh _1 s, y/l + s 2 cos v, y/\ + s 2 sin v).
Thus by the preceding lemma
w x _ ^) _ "I
AW ~ />(«) (1 +s 2 ) 2 "
This formula for Gaussian curvature in terms of x is consistent with the
formula K(u) = — l/cosh 4 w found in Example 6.1 for the parametrization x.
In fact, since s(u) = sinh u, we have
K(s(w)) = (1 +s 2 (w)) 2 = (1 +sinh 2 w) 2 = cosh 4 w '
The simple formula for K in Lemma 6.3 suggests a way to construct
surfaces of revolution with prescribed Gaussian curvature. Given a func
tion K on some interval, we first solve the differential equation h" + Kh =
for h, subject to initial conditions h(0) > and  h' (0) \ < 1. To get a
canonical parametrization we need a function g such that g + h =1.
Evidently
g{u) = f Vl  h"(t) dt
will do the job. Thus for any interval around on which the conditions
h > and  h'  < 1 both hold, we can revolve
a(u) = (g(u), h(u), 0)
around the x axis to obtain a surface of revolution whose Gaussian curva
ture is, by Lemma 6.3, precisely —h"/h = K.
6.5 Example Surfaces of revolution with constant positive curvature.
240 SHAPE OPERATORS
[Chap. V
We apply the procedure above to the constant function K = 1/c 2 The
differential equation h" + (l/c)h = has general solution
h(u) = a cos ( + 6 J
The constant b represents merely a translation of coordinates; we may
suppose that b = and a > 0. Thus the functions
gin) = f U a/ 1 _<L
h(u) = a cos 
sin  dt
c
give rise to a surface of revolution M a with constant Gaussian curvature
K = 1/c . The necessary conditions h > and  ti \ < 1 determine the
interval / to which u must be restricted. The constant c is fixed throughout
the discussion, but the constant a is at our disposal; we consider three cases.
Cose 1. a = c. Here
g(u) = / cos  dt = c sin  ,
J o c c
and h(u) = c cos (u/c). The interval i" is thus ttc/2 < u < T c/2. Since
this profile curve a(u) = (g(u), h(u)) is a semicircle, revolution about
the x axis produces the sphere 2 of radius c— except for its two points on
the x axis.
Cose 2. < a < c (Fig. 5.42). Here h is positive on the same interval
as above, and  ti \ < 1 is always true, so g is welldefined. The profile
curve u > (g(u), h(u)) still has the same length ttc/2, but it now makes
a shallower arch, which rests on the x axis at ±a*, where
4 C KV^!<
Approximate values for this elliptic integral may be found in tables,
but clearly as a shrinks from c down to 0, a* increases from c to xc/2. The
Sec. 6] SURFACES OF REVOLUTION 241
resulting surface of revolution M a — round for a = c — becomes football
shaped at first, and for a very small is a needle of length just less than xc/2.
By contrast with Case 1, the intercepts (±a*, 0, 0) cannot be added to
M a now, since this surface is actually pointed at each end. The differential
equation h," + (1/c 2 ) A = has delicately adjusted the shape of M a so
that its principal curvatures are no longer equal, but still give
K = k li k r = 1/c 2 . "
Case 3. a > c (Fig. 5.43). Here the inter
val / is reduced in size, since the expression
under the square root (in the formula for g) /^ ,i \
becomes zero at £*such that sin (t /c) = c/a ""
< 1.
Thus
Va 2  (?/
($)  *
h(t*) = a cos I — J = v a 2 — c 2 . fig. 5.43
As a increases from a = c, the resulting surface of revolution M a is at first
somewhat like the outer half of a torus, but when a is very large, it becomes
an enormous circular band whose very short profile curve is sharply curved.
(kp must be large, since K ~ I /a is small and k,Jk r = 1/c .)
A similar analysis for constant negative curvature leads to an infinite
family of surfaces of revolution with K = — 1/c 2 (Exercise 9). The simplest
of these is
6.6 Example The bugle surface B. The profile curve of B (in the xy
plane) is characterized by this geometric condition: It starts at the point
(0, c) and moves so that its tangent line always reaches the x axis after
running for distance exactly c. This curve (a tractrix) may thus be de
scribed analytically by a(u) — (u, h(u)), u > 0, where h is the solution
of the differential equation h = —h/y/c 2 — h 2 such that h(u) ^ c as
u — > 0. (The resulting surface of revolution is shown in Fig. 5.44.)
Using this differential equation, we deduce from the formulas (3) on
page 235 that the principal curvatures of B are
k =!l k  1
c ch
Thus the bugle surface (or tractroid) has constant negative curvature
K = 1/c 2 .
This surface cannot be extended across its rim (not part of B) to form a
larger surface in E 3 , since A* M (w) — > °o as u — > 0.
242
SHAPE OPERATORS
[Chap. V
FIG. 5.44
When this surface was first discovered, it seemed to be the analogue,
for K constant negative, of the sphere; it was thus called a pseudosphere.
However as we shall see in Chapter VII the true analogue of the sphere is
quite a different surface and cannot be found in E 3 .
EXERCISES
1. Find the Gaussian curvature of the surface obtained by revolving the
curve y = e~ x /2 around the x axis. Sketch this surface and indicate the
regions where K > and K < 0.
2. (Sign of Gaussian curvature). When y = f(x) is revolved about the x
axis, and K is expressed in terms of x, show that K has the same sign
( — , 0, + ) as —/" for each value of x. So K is positive on parallels through
convex intervals on the profile curve, and negative for concave intervals,
as shown Fig. 5.45. (The same result is true for an arbitrary surface of
revolution, with convexity and concavity taken relative to the axis of
revolution A.)
3. (Magnitude of Gaussian curvature). Show that
I K( u ) I — curvature k(u) of profile curve a at a(u),
 k r (u)  = h(u)  cos <p(u) , where <p(u) is the slope angle of the
profile curve at a(u),
hence that
 K  = ich  cos <p  .
Sec. 6]
SURFACES OF REVOLUTION
243
K >
K >
K <
4
FIG. 5.45
tf >
K =
X <0
FIG. 5.46
4. An elliptical torus M is obtained by revolving an ellipse
(x  R) 2 /a + 2/V& 2 = 1
about the y axis (R > a). Find a parametrization for M and compute
its Gaussian curvature. (Check your answer by setting a = b = r.)
5. If r = *\/x 2 + y 2 > is the usual polarcoordinate function on the xy
plane, and / is a differentiate function, show that M : z = f(r) is a
surface of revolution, and that its Gaussian curvature (expressed in
terms of r) is
K ^ f(r)f''(r)
r(l +/'(r) 2 ) 2 "
6. Find the Gaussian curvature of the surface M: z = e~ r /2 . Sketch this
surface, indicating the regions where K > and K < 0.
7. Prove that a flat surface of revolution is part of a cone or a cylinder.
8. Let M be the surface obtained by revolving one arch ( — k < t < ir)
of the cycloid 7(<) = (t + sin t, 1 f cos t) about the x axis.
(a) Compute K relative to the usual parametrization of M.
(b) Find the function h giving the height of a in terms of its arc length
(measured from the top of the arch). Computed = —h"/h.
{Hint: Use half angles.)
(c) Show that the results in (a) and (b) are consistent.
244 SHAPE OPERATORS [Chap. V
9. (Surfaces of revolution with constant negative curvature K = — 1/c 2 ).
As in the case K = 1/c 2 , there is a family of such surfaces, divided into
two subfamilies by a special surface. The solutions of
h"  h/c 2 =
giving, by canonical parametrization, essentially all these surfaces are:
(a) h(u) = a sinh (u/c), < a < c, u > 0. Show that the profile curve
a (u) = (g(u), h(u)) leaves the origi n with s lope a/\/ c 2 — a 2 and
rises toward a maximum height of \/c 2 — a 2 . Sketch the resulting
surface of revolution M a for a small value of the parameter a, and
for a near c.
(b) h(u) = b cosh (u/c), b > 0. Show that the profile curve a rises
symmetrically (for ±w) toward a maximum height of s/c 2 + b 2 .
Sketch the resulting surface M b for b small and for b large.
(c) h(u) = ce u,c , u < 0. This surface B is merely a mirror image of
the surface B gotten from
h(u) = ce , u > 0.
Show that B is, in fact, the bugle surface (Example 6.6). How does
the surface B separate the two subfamilies, that is, for which values
of a and b do M a and M b resemble B? (See Fig. 5.46 where M a and
M b have been translated along the axis of revolution.)
7 Summary
The shape operator S of a surface M in E 3 measures the rate of change of
a unit normal U in any direction on M. If we imagine U as the "first
derivative" of M, then S is the "second derivative." But the shape opera
tor is also an algebraic object consisting of linear operators on the tangent
planes of M . And it is by an algebraic analysis of S that we have been led
to the main geometric invariants of a surface in E 3 : its principal curvatures
and directions, and its Gaussian and mean curvatures.
CHAPTER
VI
Geometry of Surfaces in E 3
Now that we know how to measure the shape of a surface M in E , the
next step is to see how the shape of M is related to itsother properties. Near
each point of M, the Gaussian curvature has a strong influence on shape
(Remark 3.3 of Chapter V), but we are now interested in the situation in
the large — over the whole extent of M. For example, what can be said about
the shape of M if it is compact, or flat, or both?
Almost 150 years ago Gauss raised a question that led to a new and
deeper understanding of what geometry is: How much of the geometry of
a surface in E 3 is independent of its shape? At first glance this seems a strange
question — what can we possibly say about a sphere, for example, if we
ignore the fact that it is round? To get some grip on Gauss's question, let
us imagine that the surface McE s has inhabitants who are unaware of
the space outside their surface, and thus have no conception of its shape in
E 3 . Nevertheless, they will still be able to measure "the distance from place
to place in M and find the area of regions in M. In this chapter and the
next we shall see that in fact they can construct an "intrinsic geometry"
for M that is richer and no less interesting than the familiar Euclidean
geometry of the plane E 2 .
1 The Fundamental Equations
To study the geometry of a surface M in E we shall apply the Cartan
methods outlined in Chapter II. As with the Frenet theory of a curve in
E 3 , this requires that we put frames on M, and examine their rates of change
along M . Formally, a Euclidean frame field on M C E consists of three
245
246 GEOMETRY OF SURFACES IN E 3 [Chap. VI
E,
FIG. 6.1
Euclidean vector fields (Definition 3.7, Chapter IV) that are orthonormal
at each point. Such a frame field can be fitted to its surface as follows.
1 .1 Definition An adapted frame field E h E 2 , E 3 on a region in M CZ E 3
is a Euclidean frame field such that E 3 is always normal to M (hence E\
and E 2 are tangent to M) (Fig. 6.1).
Thus the normal vector field denoted by U in the preceding chapter now
becomes E 3 . For brevity we shall refer to an adapted frame field "on M,"
but the actual domain of definition is in general only some region in M,
since an adapted frame field need not exist on all of M.
1.2 Lemma There is an adapted frame field on a region in M a E 3 if
and only if is orientable and there exists a nonvanishing tangent vector
field on 0.
Proof. This condition is certainly necessary, since E 3 orients 0, and E x
and E 2 are unit tangent vector fields. To show that it is sufficient, let be
oriented by a unit normal vector field U, and let V be a tangent vector
field that does not vanish on 0. But then it is easy to see that
Ex = Sj , E 2 = U XE h E 3 = U
is an adapted frame field on 0. 
1.3 Example Adapted frame fields.
(1) Cylinder M: x 2 + y = r 2 . The gradient of g = x 2 + y 2 leads to the
unit normal vector field E 3 = (xUi + yU 2 )/r. Obviously the unit vector
field U% is tangent to M at each point. Setting E 2 = U 3 X E 3 , we then
get the adapted frame field
Ey = U 3
yVx + xU 2
r
xUi + yU 2
Hj 3 =
r
on the whole cylinder M (Fig. 6.2).
Sec. 1]
THE FUNDAMENTAL EQUATIONS
247
(2) Sphere 2: x + y 2 + z = r\ The outward
unit normal
E% =
xUx + yU 2 + zU*
is defined on all of 2, but as we shall see in
Chapter VII, every tangent vector field on 2
must vanish somewhere. For example, the "due
east" vector field V = — yUx + xU 2 is zero at the
the north and south poles (0,0, ±r). Thus the
adapted frame field
E x =
V
M
E,
K E t
FIG. 6.2
Ei = E% X E\
Es =
xUx + yU 2 + zU 3
(Fig. 6.3) is defined on the region in 2 gotten by deleting the north and
south poles.
Lemma 1.2 implies in particular that there is an adapted frame field on
the image x(£>) of any patch in M; thus such fields exist locally on any
surface in E .
Now we shall bring the connection equations (Theorem 7.2 of Chapter
II ) to bear on the study of a surface M in E 3 . Let E h E 2 , E 3 be an adapted
frame field on M. By moving each frame Ei(p), E 2 (p), E 3 (p) over a short
FIG. 6.3
248 GEOMETRY OF SURFACES IN E 3 [Chap. VI
interval on the normal line at each point p, we can extend the given frame
field to one defined on an open set in E 3 . Thus the connection equations
are available for use. We shall apply them only to vectors v tangent to M. In
particular, the connection forms w»y become 1forms on M in the sense of
Section 4 of Chapter IV. Thus we have
1.4 Theorem If Ei,E 2 ,E 3 is an adapted frame field on M C E 3 , and v
is tangent to M at p, then
V^ = £ Wt y(v)^(p) (1 ^ i ^ 3).
The usual interpretation of the connection forms may be read from these
equations, and it bears repetition: o> t j(v) is the initial rate at which Ei ro
tates toward E, as p moves in the v direction. Since E 3 is a unit normal vector
field on M, the shape operator of M can be described by connection forms.
1.5 Corollary Let S be the shape operator gotten from E 3 , where
Ei, Ei, E 3 is an adapted frame field on M C E 3 . Then for each tangent
vector v to M at p,
£(v) = W U (v).Ei(p) + C028(v)# 2 (p)
Proof. By definition, S(\) = — V V E 3 . Thus the connection equation for
i — 3 gives the result, since the connection form co = (tan) is skewsym
metric: Uij = — C0/i. 
In addition to its connection forms, the adapted frame field E lt E 2 , E 3
also has dual 1forms 8 i} 2 , 03 (Definition 8.1 of Chapter II) which give
the coordinates t (v) = v£j(p), of any tangent vector \ p with respect to
the frame i?i(p), ^(p), ^(p) As with the connection forms, the dual
forms will be applied only to vectors tangent to M, so they become forms
on M. This restriction is fatal to 6 S , for if v is tangent to M, it is orthogonal
to E 3 , so 3 (v) = v£i(p) = 0. Thus 3 is identically zero on M.
Because of the skewsymmetry of the connection form, we are left with
essentially only five 1forms:
0i, 2 provide a dual description of the tangent vector fields E u E 2
ooi2 gives the rate of rotation of Ei, E 2
^13, ^23 describe the shape operator derived from E 3
1.6 Example The sphere. Consider the adapted frame field E lf E 2 , E 3
defined on the (doubly punctured) sphere 2 in Example 1.3. By extending
this frame field to an open set of E we get the spherical frame field given in
Example 6.2 of Chapter II, provided the indices of the latter are shifted by
Sec. 1] THE FUNDAMENTAL EQUATIONS 249
1 — ► 3, 2 — ► 1, 3 — » 2. Thus, in terms of the spherical coordinate functions,
Example 8.4 of Chapter II gives
6i = r cos <p d& 2 = r d<p
cow = sin <p d& coi3 = — cos $? d# 023 = —dtp
Because all forms (including functions) are now restricted to the surface
2, the spherical coordinate function p has become a constant: the radius r
of the sphere.
In general the forms associated with an adapted frame field obey the
following remarkable set of equations.
1.7 Theorem If E ly E 2 , E z is an adapted frame field onlc E 3 , then its
dual forms and connection forms on M satisfy:
\dd\ = OJi2 A 2
(1)1 First structural equations
Id02 = U21 a 6\
(2) a>3i a 0i + 0332 a 02 — Symmetry equation
(3 ) do3i2 = o>i3 a o> 3 2 Gauss equation
[do>i3 = OJi2 A C023
(4) Codazzi equations
do>23 = «21 A OJ]3
Proof. We merely apply the structural equations in Theorem 8.3 of
Chapter II. The first structural equation
ddi — zl Ma a dj
i
yields (1) and (2) above. In fact, for i = 1, 2, we get (1), since 3 =
on the surface M. But 3 = implies d6 3 = 0, so for i = 3 we get (2).
Then the second structural equation yields the Gauss (3) and Codazzi
(4) equations. I
Because connection forms are skew symmetric, and a wedge product of
1 forms satisfies <j> a \p = —\f/ a <f>, the fundamental equations above can
be rewritten in a variety of equivalent ways. However, we shall stick to
the index pattern used above, which, on the whole, seems the easiest to
remember.
We emphasize that the forms introduced in this section describe, not the
surface M directly, but only the particular adapted frame field E lf E 2 , E z
from which they are derived : A different choice of the frame field will pro
duce different forms. Nevertheless the six fundamental equations in Theorem
1.7 contain a tremendous amount of information about the surface M c E ,
and we shall call on each in turn as we come to a geometric situation that
250
GEOMETRY OF SURFACES IN E 3
[Chap. VI
it governs. For example, since o>i 3 and 0023 describe the shape operator of M,
the Codazzi equations (4) express the rate at which the shape of M is
changing from one point to another.
The first of the following exercises shows how the Cartan approach
automatically singles out the three types of curves considered in Chapter
V, Section 5.
EXERCISES
1. Let a be a unitspeed curve in M C E 3 . If Ei, E 2 , E 3 is an adapted frame
field such that E x restricted to a is its unit tangent T, show that
(a) a is a geodesic of M if and only if a>i 2 (T) = 0.
(b) If E 3 = EtX E 2 , then
g = un(T), k = Wl3 (T), t = u> 23 {T)
where g, k, and t are the functions defined in Ex. 7 of V.5. {Hint: If T =
Ei along a, then E/ = V El Ei along a. )
2. {Sphere). For the frame field in Example 1.6:
(a) Verify the fundamental equations (Theorem 1.7).
(b ) Deduce from the formulas for 0i and 2 that
E x [#\ = l/r cos <P E } [<p] =
E 2 [d] = Et[<p] = l/r
(c) Use Corollary 1.5 to find the shape operator S of the sphere.
3. {Torus). Let E u E 2 , E 3 be the adapted frame field on the torus T (radii
R > r) such that E 2 is tangent to meridians and E x is tangent to paral
lels, as in Fig. 6.4. Use the toroidal frame field in E 3 to get
0i = (R + r cos <p) d#
2 = r d<p
CJ12 = sin <p d#
o)i3 = — cos <p d&
C0 2 3 = —dip
Check the fundamental
equations for these forms.
4. {Continuation). By the
fig. 6.4 methods of this section,
Sec. 2] FORM COMPUTATIONS 251
compute S(Ei) and S(E 2 ) for the frame field above. Deduce that
meridians and parallels are principal curves, and find the principal cur
vature functions. (Compare Example 6.1 of Chapter V, where the unit
normal is "inward.")
5. Use the cylindrical frame field in E 3 (Example 6.2 of Chapter II) to
compute the shape operator of the cylinder M: x + y = r 2 .
6. Give a new proof that shape operators are symmetric by using the
symmetry equation (Theorem 1.7).
2 Form Computations
From now on, our study of the geometry of surfaces will be carried on
mostly in terms of differential forms, so the reader may wish to look back
over their general properties in Sections 4 and 5 of Chapter IV. Increas
ingly we shall tend to compare M with the Euclidean plane E 2 . Thus if
Ei, E 2 , E z is an adapted frame field onMc E 3 , we say that E lf E 2 con
stitutes a tangent frame field on M. Any tangent vector field V on M may
be expressed in terms of Ei and E 2 by the orthonormal expansion
V = V>EiEi+ V'E 2 E 2
To show that two forms are equal, we do not have to check their values on
all tangent vectors, but only on the "basis" vector fields E x , E 2 . (See the
remarks preceding Example 4.7 of Chapter IV). Explicitly: lforms <t> and
yp are equal if and only if
<j>(Ei) = +(Ei) and <t>(E 2 ) = +(E 2 );
2forms n and v are equal if and only if
niE^E^ = v(Ei,E 2 ).
The dual forms d u 6 2 are, as we have emphasized, merely another descrip
tion of the tangent frame field E x , E 2 ; they are completely characterized by
the equations
e t (Ei) = 8 {j (1 £i,j ^2).
These forms provided a "basis" for the forms on M (or, strictly speaking,
on the region of definition of E u E 2 ).
2.1 Lemma (The Basis Formulas) Let di, 6 2 be the dual lforms of
Ei, E 2 on M. If is a 1form and n a 2form, then
(1) $ = <f>(Ei) e x +4>(E 2 ) 2
(2) M = p(E u E 2 ) d x a 2
252 GEOMETRY OF SURFACES IN E 3 [Chap. VI
Proof. Apply the equality criteria above, observing for (2) that by
definition of the wedge product,
(0i a 6 2 )(Ex,E 2 ) = d 1 (E 1 )6 2 (E 2 )  6 1 (E 2 )d 2 (E 1 )
= 11  00 = 1. I
Assuming throughout that the forms X , 2 , o> n , wu, w 23 derive as in Section
1 from an adapted frame field Ex, E 2 , E 3 on a region in M, let us see what
some of the concepts introduced in Chapter V look like when expressed in
terms of forms. We begin with the analogue of Lemma 3.4 of Chapter V.
2.2 Lemma
(1) 0>i3 A C0 2 3 = K6i A 2 .
(2) WW A 2 f X A a>23 = 2#0! A 6 2 .
Proof. To apply the definitions K = det S, 2H = trace S, we shall find
the matrix of S with respect to E x and E 2 . As in Corollary 1.5 the connection
equations give
S(Ex) = V El E s = o} 31 (Ex)Ex  u Z2 {Ex)E 2
S(E 2 ) = V E2 E 3 = u> n (E 2 )Ex  u Z2 (E 2 )E 2 .
Thus the matrix of S is
(wx%(Ex) uisiExY
\wxi{E 2 ) oo 2 z(E 2 ) t
Now, by the second formula in Lemma 2.1, what we must show is that
(cow a comX^i, E z ) = K and (wu a 2 f di a wa)(^i, ^2), = 2H.
But
(ww A u n ){Ex,E 2 ) = oixz{Ex)oi 2i {E 2 )  Uiu{E 2 )o^{Ex)
= determinant of matrix of S = det S = K
and a similar computation gives the trace formula. 
Comparing the first formula above with the Gauss equation (3) in
Theorem 1.7, we get
2.3 Corollary dw l2 = —Kdx a 2 .
We shall call this the second structural equation,] and derive from it a
new interpretation of Gaussian curvature: u n measures the rate of rotation
of the tangent frame field E u E 2 —a,nd since K determines the exterior
derivative dw 12 , it becomes a kind of "second derivative" of E u E 2 .
t This equation will be shown to be the analogue for M of the second struc
tural equation (Theorem 8.3 of Chapter II) for E 3 .
Sec. 2] FORM COMPUTATIONS 253
For example, on a sphere 2 of radius r, the formulas in Example 1.6 give
di a 2 = r cos <p d& d<p = — r 2 cos <p d<p d&.
But
do)^ = d(sin<pd&) = d(sin.(p) a d& = cos <p d*pdd.
Thus the second structural equation gives the expected result, K = 1/r 2 .
This new description of curvature may be rewritten in still another way.
2.4 Corollary K = # 2 [a>i 2 (#i)]  #i[» u (tf 2 )]  ^{Exf  ^(E 2 )\
Proof. By Lemma 2.1, we have
Wl2 = fl6i + / 2 2 ,
where
fi = un(Ei) fori = 1, 2.
Then
dun = rf/l A 0! + rf/ 2 A 2 + /i d0 x + / 2 d0 2
= d/ X A 0! + rf/ 2 A 2 + /lCO i2 A 2 + /2C02! A X
where we have used the first structural equations (Theorem 1.7). Now
apply this formula to E u E 2 . Since 0;(2£y) = 5 t y, we get
douiE^Ei) = dME 2 ) +df i (E 1 ) +fi<» a (Ei)  fa^iE*)
Hence, using the previous corollary,
— K= — E 2 [fi] + Ei[fi] \ f\o)n(Ei) +/ 2 c»)i 2 (E 2 )
which, by definition of /i and / 2 , is the required result. 
For instance, from Example 1.6 we readily compute that
uniEy) =  tan <p and o)\ 2 {E 2 ) = 0.
r
Thus f or the sphere, the formula above yields
2
K = EA tan <p — [  tan <p J = —
since by Exercise 2 of Section 1 we have
— tan tp _ 1
£"2 [tan <p] = sec 2 <p E 2 [<p] =
T
We emphasized in Section 1 that, in general, adapted frame fields on
M (Z E give only indirect information about M. If such a frame field is to
254 GEOMETRY OF SURFACES IN E 3 [Chap. VI
give direct geometric information, it must be derived in some natural way
from the geometry of M itself, as was the case with the Frenet frame field
of a curve. There is a way to do this :
2.5 Definition A principal frame field on M C E is an adapted frame
field Ei, E 2 , E 3 such that at each point E x and E 2 are principal vectors of M.
So long as its domain of definition contains no umbilics, a principal
frame field is uniquely determined — except for changes of sign — by the
two principal directions at each point.
Occasionally it may be possible to get a principal frame field on an entire
surface. For example, on a surface of revolution, we can take E\ tangent to
meridians, E 2 tangent to parallels. In general, however, about the best we
can do is as follows.
2.6 Lemma If p is a nonumbilic point of I c E 3 , then there exists a
principle frame field on some neighborhood of p in M.
Proof. Let Fi, F 2 , Fz be an arbitrary adapted frame field on a neighbor
hood 91 of p. Since p is not umbilic, we can assume (by rotating Fi, F 2
if necessary) that Fi(p) and Fi (p) are not principal vectors at p. By
hypothesis fci(p) ?± fc 2 (p); hence by continuity ki and k 2 remain distinct
near p. On a small enough neighborhood 31 of p, all these conditions are
thus in force.
Let Sij be the matrix of $ with respect to F u Fi. It is now just a standard
problem in linear algebra to compute — simultaneously at all points of 91 — 
characteristic vectors of S, that is, principal vectors of M. In fact, at each
point the tangent vector fields
7i = S 12 F t + (h  S U )F 2
V 2 = (k 2 — $22)^1 + S\ 2 F 2
give characteristic vectors of S. (This can be checked by a direct computa
tion if one does not care to appeal to linear algebra.) Furthermore, the
function $12 = S(Fi)'F 2 is never zero on our selected neighborhood 91,
so  Vi  and  V 2  are never zero. Thus the vector fields
El =  J7 El =
.. Vi 11 11 y 2 11
consist only of principal vectors, so E u E 2 , E 3 = F 3 is a principal frame
field on 91. 
If Ei, Ei, Ez is a principal frame field on M, then the vector fields Ei and
E 2 consist of characteristic vectors of the shape operator derived from E z .
Thus we can label the principal curvature functions so that S (Ei ) = k t Ei
and S(E 2 ) = k 2 E 2 . Comparison with Corollary 1.5 then yields
Sec. 2] FORM COMPUTATIONS 255
CtflsCEl) = fcl 03 U (E2) =
um{Ex) = 0023(^2) = fa
Thus the basis formula (1) in Lemma 2.1 gives
C0i3 = fadl 0>23 = fa&2 ( )
This leads to an interesting version of the Codazzi equations.
2.7 Theorem If E u E 2 , E s is a principal frame field on M C E 3 , then
EM = (fa ~ fa) u l2 (E 2 )
E 2 [fa] = (fa  fa)o, 12 (E l )
Proof. The Codazzi equations (Theorem 1.7) read
UOliz = Wl2 A OJ23 0,0323 = W21 ^ W13.
The proof is now an exercise in the calculus of forms as discussed in Chapter
IV, Section 4. Substituting from (*), above, in the first of these equations,
we get
d(fadi) = CO12 a fad 2 ;
hence
dfa a di \ fa dd]_ = fc 2 aji2 a 2
If we substitute the structural equation dB\ = a>i 2 a 6 2 , this becomes
dfa A 6 1 = (&2 — fa) CO12 A 2
Now apply these 2forms to the pair of vector fields E if E 2 to obtain
 dfa(E 2 ) = (fa  fa) w M C&i)  0,
hence
E 2 [fa] = dfa(E 2 ) = (fa  fa) ««(#!).
The other required equation derives in the same way from the Codazzi
equation dutx = 0021 a W13. I
Note that for a principal frame field, uu(v) tells how the principal
directions are changing in the v direction.
EXERCISES
1. Verify the Codazzi equations (Theorem 2.7) for the principal frame
field on the torus given in Exercise 3 of Section 1. (Hint:
12 
gV[f)  fV[g]
256 GEOMETRY OF SURFACES IN E 3 [Chap. VI
the derivatives of # and <p with respect to E x and E 2 may be found from
the formulas for X and 2 in this exercise.)
2. If E h E 2 , E 3 is an adapted frame field on M with E x • E 2 X E 3 = 1, let
Mp) = P • #«(p) for i = 1, 2, 3.
In particular, hz is the support function h of M defined on page 218.
Show that
dhi = 0i + (dwhi + W13/13
<//&2 = 02 + 0)21^1 + G>23^3
(flm£: Strictly speaking, h { = X > E iy where X is the remarkable vector
field such that V V X = V used also on page 218.)
There is a rough rule for computing exterior derivatives of 1 forms in
terms of an adapted frame field: Express the form in terms of 0i and 2
(or perhaps &»,,•); then apply d and use the fundamental equations. The
proof of Theorem 2.7 is one example; another is as follows.
3. (Continuation).
(a) If ^ is the 1form such that ^(v) = p • v X ^(p), show that
# = 2(1 + hH) X a 2 ,
where h is the support function.
(b) If f is the 1form such that f (v) = ^(S(v)), show that
rff = 2(H + hK) 0i a 2 .
(Hint: \f/ = hi 2 — /i 2 0i, and f has an analogous expression.)
3 Some Global Theorems
We have claimed all along that the shape operator S is the analogue for
a surface M of the curvature and torsion of a curve in E 3 . Simple hypotheses
on k and t singled out some important special types of curves. Let us now
see what can be done with S in the case of surfaces. (Recall that we are
dealing exclusively with connected surfaces. )
3.1 Theorem If its shape operator is identically zero, then M is (part
of ) a plane in E 3 .
Proof. The scheme used is analogous to that of Corollary 3.5 of Chapter
II. By definition of the shape operator, S = means that any unit normal
vector field E z on M is Euclidean parallel; thus it may be identified with a
point of E 3 . Fix a point p of M. We shall show that M lies in the plane
Sec. 3]
SOME GLOBAL THEOREMS
257
FIG. 6.5
through p orthogonal to E 3 . If q is an arbitrary point of M , then since M
is connected, there is a curve a in M from a (0) = p to a (1 ) = q. Consider
the function
Now
df
dt
f(t) = («(*) p) .E».
a*E 3 = and /(0) = 0;
hence / is identically zero. In particular,
/(I) = (qp) E 3 = 0,
so every point q of M is in the required plane (Fig. 6.5).
I
We saw in Chapter V, Section 3 that requiring a single point p of M C E 3
to be planar (&i = k 2 = 0, or equivalently *S = 0) produces no significant
effect on the shape of M near p. But the result above shows that if every
point is planar, then M is, in fact, part of a plane.
Perhaps the next simplest hypothesis on a surface M in E 3 is that at
each point p, the shape operator is merely scalar multiplication by some
number — which a priori may depend on p. This means that M is all
umbilic, that is, consists entirely of umbilic points.
3.2 Lemma If M is an allumbilic surface in E 3 , then M has constant
Gaussian curvature K ^ 0.
Proof. Let E x , E 2 , E 3 be an adapted frame field on some region in M.
Since M is allumbilic, the principal curvature functions on are equal,
Jd = kt = k, and furthermore E u E 2 , E 3 is actually a principal frame field
(since every direction on M is principal). Thus we can apply Theorem
2.7 to conclude that E x [k] = E 2 [k] = 0. Alternatively we may write
dkiEi) = dk(E 2 ) = 0,
258 GEOMETRY OF SURFACES IN E 3 [Chap. VI
so by Lemma 2.1, dk = on 0. But K = kik 2 = k 2 , so dK = Ik dk =
on 0. Since every point of M is in such a region 0, we conclude that dK =
on all of M. It follows by an earlier exercise that K is constant. 
3.3 Theorem If M d E 3 is allumbilic and K > 0, then M is part of
a sphere in E 3 of radius \/\/K.
Proof. (This time the scheme of proof is analogous to Lemma 3.6 of
Chapter II.) Pick at random a point p in M and a unit normal vector
^(p) to\M at p. We shall prove that the point
c = p + ^) Ea(p)
is equidistant from every point of M. (Here A;(p) = fci(p) = fc 2 (p) is the
principal curvature corresponding to 2£ 3 (p).)
Now let q be any point of M, and let a be a curve segment in M from
a(0) = ptoa(l) = q. Extend E^p) to a unit normal vector field E 3 on
a, as shown in Fig. 6.6, and consider the curve
7 = a + r Ez in E .
k
Here we understand that the principal curvature function k derives
from E 3 , thus k is continuous. But K = k 2 and by the preceding lemma,
K is constant, so k is constant. Thus
7 = a + — Ez .
But
E 3 ' = S(a) =  ka,
E,(p) E t
Sec. 3] SOME GLOBAL THEOREMS 259
since by the allumbilic hypothesis, S is scalar multiplication by k. Thus
7  a + t (~ka)  0,
so the curve 7 must be constant. In particular
c = 7(0) = 7(D = q+i# 3 (q)
so <l(c, q) = 1/ I k I for every point q of M. Since K = kik 2 = k 2 , we
have shown that M is contained in the sphere of center c and radius
i/Vk. I
Using all three of the preceding results, we conclude that a surface M
in E is allumbilic if and only if M is part of a plane or a sphere.
3.4 Corollary A compact allumbilic surface M in E is an entire sphere.
Proof. By the preceding remark, we deduce from Theorem 7.6, Chapter IV,
that M must be an entire plane or sphere. The former is impossible, since
M is — by hypothesis — compact, but planes are not. 
Gaussian curvature was used in the preceding results mostly because
it is welldefined and differentiable on all of M, and is thus easier to work
with than principal curvatures.
We now turn to a more serious examination of the Gaussian curvature
K of a surface M (Z E 3 .
3.5 Theorem On every compact surface M in E 3 , there is a point at
which the Gaussian curvature K is strictly positive.
Proof. Consider the realvalued function/ on M such that/(p) =  p  2 .
Thus in terms of the natural coordinates of E 3 , / = ^2 %?• Now / is differ
entiable, hence continuous, and M is compact. Thus by Lemma 7.3 of
Chapter IV, / takes on its maximum at some point m of M . Since / meas
ures the square of the distance to the origin, m is simply a point of M at
maximum distance r =  m  > from the origin. Intuitively it is clear
that M is tangent at p to the sphere 2 of radius r — and that M lies inside
S, hence is more curved than 2 (Fig. 6.7). Thus we would expect that
K(m) > 1/r 2 > 0. Let us prove this inequality.
Given any unit tangent vector u to M at the maximum point m, pick
a unitspeed curve a in M such that a(0) = m, a (0) = u. It follows
from the derivation of m that the composite function f(a) also has a
maximum at t = 0. Thus
(/a)(0)=0,  2 (/«)(0) ^0. (1)
260 GEOMETRY OF SURFACES IN E 3 [Chap. VI
FIG. 6.7
j
But /(a) = a*a, so r(fa) = 2a • a. Evaluating at t = 0, we find
= ^M. (0) = 2a(0).«'(0) = 2m.u.
at
Since u was any unit tangent vector to M at m, this means that m (con
sidered as a vector) is normal to M at m.
Differentiating again, we get
d (fa) n ' ' s n tf
eft 2
By (1), aU = this yields
> u • u + m • a" (0)
(2)
= 1 + m .a"(0).
The discussion above shows that m/r may be considered as a unit normal
vector to M at m as shown in Fig. 6.8. Thus (m/r)»a" is precisely the
normal curvature k(u) of M in the u direction, and it follows from (2)
that k(u) ^ — 1/r. In particular, both principal curvatures satisfy this
inequality, so
K(m)^ 2 >0. 
Thus there are no compact surfaces in E with K ^ 0.
Maintaining the hypothesis of compactness, we consider the effect of
requiring that Gaussian curvature be constant. Theorem 3.5 shows that
the only possibility is K > 0. Spheres are obvious examples of compact
surfaces in E 3 with constant positive Gaussian curvature. It is one of the
most remarkable facts of surface theory that they are the only such sur
faces. To prove this we need a rather deep preliminary result.
Sec. 3]
SOME GLOBAL THEOREMS
m/r
261
FIG. 6.8
3.6 Lemma (Hilbert) Let m be a point of McE such that
(1) ki has a local maximum at m.
(2) fc 2 has a local minimum at m.
(3) fcx(m) > k(m).
ThenK(m) ^ 0.
For example, it is easy to see that these hypotheses hold at any point
on the inner equator of a torus or on the minimal circle (x = 0) of the
catenoid. And K is, in fact, negative in both these examples.
To convert hypotheses (1) and (2) into usable form in the proof that
follows, we recall some facts about maxima and minima. If / is a (differ
entiable) function on a surface M and V is a tangent vector field, then the
first derivative V\f] is again a function on M. Thus we can apply V again
to obtain the second derivative F[F[/]] = FFf/]. A straightforward com
putation shows that if / has a local maximum at a point m, then the ana
logues of the usual conditions in elementary calculus hold, namely
V[f] = 0, VV[f] SO at m.
For a local minimum, of course, the inequality is reversed.
Proof. Since fci(m) > /^(m), m is not umbilic, hence by Lemma 2.6
there exists a principal frame field E it E 2 , E 3 on a neighborhood of m in
M. By the remark above, the hypotheses of minimality and maximality
at m imply in particular
EM = Etlh] = atm (1)
and
EiEM ^ and E^lh] ^0 at m. (2)
262 GEOMETRY OF SURFACES IN E 3 [Chap. VI
Now we use the Codazzi equations (Theorem 2.7). From (1) it follows
that
oin(Ei) = con(E 2 ) =0 at m
since fci — k 2 5* at m, thus by Corollary 2.4
K = EJLwniEi)]  EtlaniEi)] at m. (3)
Applying Ei to the first Codazzi equation in Theorem 2.7 yields
EMh] = CEitfci]  E^h]) <**(E 2 ) + (fci  A*) #i[« B (#,)].
But at the special point m, we have o>n(E 2 ) = and fci — k 2 > 0; hence
from (2 ) we deduce
Ei[an(E 2 )] ^ atm. (4)
A similar argument starting from the second Codazzi equation gives
EtiuniEi)] S atm. (5)
Using (4) and (5) in the expression (3) for the Gaussian curvature at m,
we conclude that K(m) ^ 0. I
3.7 Theorem (Liebmann) If M is a compact surface in E with constant
Gaussian curvature K, then M is a sphere of radius l/\/K. (Theorem 3.5
implies K is positive.)
Proof. We do not know that M is orientable, so principal curvature
functions are not available on all of M. Nevertheless, the function
H 2  K = (h fc 2 ) 2 /4
is welldefined and continuous on all of M, since squaring eliminates
ambiguity of sign. Because M is compact, the function H 2 — K ^ has
a maximum point m. Now if H 2 — K is zero at m, it is identically zero;
thus M is allumbilic and Corollary 3.4 gives the required result.
So what we must show is that H 2 — K cannot be positive at m. Assume
it is; then m is not umbilic and by suitably orienting a neighborhood 91
of m we can arrange that h > k 2 > on 91 (since K > 0). Then ki — k?
has a maximum at m, since (fci — k 2 ) does. Since K = kik 2 is constant,
it follows that ki has a local maximum at m and k 2 has a local minimum.
But now we can apply Hilbert's lemma to obtain the contradiction
K(m) ^0. I
Liebmann's theorem is false if the compactness hypothesis is omitted,
for we saw in Chapter V, Section 6, that there are many nonspherical
surfaces in E 3 with constant (positive) curvature. Both Theorem 3.5 and
Liebmann's theorem depend on the fundamental topological fact (Lemma
Sec. 4] ISOMETRIES AND LOCAL ISOMETRIES 263
7.3 of Chapter IV) that a continuous real valued function on a compact
surface has a maximum. More advanced topological methods are required
for a full investigation of the influence of Gaussian curvature on the shape
of surfaces in E 3 . For example, one might ask what the situation is for
constant curvature surfaces when compactness is weakened to closure
in E 3 . f Here are the answers :
A closed surface M d E 3 with K constant positive is compact — hence
by Liebmann's theorem it is a sphere.
A closed surface M C E 3 with K = is a generalized cylinder (Massey).
There are no closed surfaces in E 3 with K constant negative (Hilbert).
We shall prove the first result in Chapter VII. Proofs of the last two may
be found in Hicks [5] and Willmore [3], respectively.
EXERCISES
1. If M is a flat minimal surface, prove that M is part of a plane.
2. Flat surfaces in E 3 can be bent only along straight lines: If fa = 0, but fa
is never zero, show that the principal curves of fa are straightline seg
ments in E 3 . (Hint: Use Ex. 1 of VI. 1.)
This is the starting point for the proof of Massey's theorem.
3. Let I c E J be a compact orientable surface with K > 0. If M has
constant mean curvature, show that M is a sphere.
4. Prove that in a region free of umbilics there are exactly two principal
curves (ignoring different parametrizations) through each point, these
crossing orthogonally. (Hint: Use Ex. 7 of IV.8.)
5. If the principal curvatures of a surface IcE 3 are constant, show that
M is part of either a plane, a sphere, or a circular cylinder. (Hint: In
the nontrivial case fa ?* fa, assume there is an adapted frame field on
all of M, and show that, say, fa = 0.)
4 Isometries and Local Isometries
We referred earlier to properties of a surface M in E that could be dis
covered by inhabitants of M unaware of the space outside their surface.
We asserted that the inhabitants of M could determine the distance in M
t This condition is defined in Exercise 10 of Chapter IV, Section 7. Roughly
speaking, it means that M has no edges or rims. For surfaces in E 3 it is equivalent
to the intrinsic property of completeness (Definition 4.4, of Chapter VII).
264
GEOMETRY OF SURFACES IN E 3
[Chap. VI
between any two points, just as the distance along the surface of the earth
is found by its inhabitants. The mathematical formulation is as follows.
4.1 Definition If p and q are points of M c E 3 , consider the collection
of all curve segments a in M from p to q. The intrinsic distance p(p, q)
from p to q in M is the greatest lower bound of the lengths L(a) of these
curve segments.
There need not be a curve a whose length is exactly />(p, q) (see Exercise
3). The intrinsic distance p(p, q) will generally be greater than the straight
line Euclidean distance d(p, q), since the curves a are obliged to stay in
M (Fig. 6.9).
On the surface of the earth (sphere of radius ca. 4000 miles) it is, of
course, intrinsic distance that is of practical interest. One says, for example,
that it is 12,500 miles from the north pole to the south pole — the Euclidean
distance through the center of the earth is only 8000.
We saw in Chapter III how Euclidean geometry is based on the notion
of isometry, a distancepreserving mapping. For surfaces in M we shall
prove the distancepreserving property and use its infinitesimal form
(Corollary 2.2 of Chapter III) as definition.
4.2 Definition An isometry F: M — ► M of surfaces in E 3 is a onetoone
mapping of M onto M that preserves dot products of tangent vectors.
Explicitly, if F* is the derivative map of F, then
F*(v)»F*(w) = vw
for any pair of tangent vectors v, w to M.
FIG. 6.9
Sec. 4] ISOMETRIES AND LOCAL ISOMETRIES 265
If F* preserves dot products, then it also preserves lengths of tangent
vectors. It follows that an isometry is a regular mapping (Chapter IV,
Section 5), for if F* (v) = 0, then
v = F«(y) =0,
hence v = 0. Thus by the remarks following Theorem 5.4 of Chapter IV,
an isometry F: M — ► M is in particular a diffeomorphism, that is, has an
inverse mapping F 1 : M — > M.
4.3 Theorem Isometries preserve intrinsic distance: if F: M — > M is
an isometry of surfaces in E 3 , then
p(p,q) =iJ(F(p),F(q))
for any two points p, q in M.
(Here p and p are the intrinsic distance functions of M and M respec
tively.)
Proof. First note that isometries preserve the speed and length of curves.
The proof is just like the Euclidean case: If a is a curve segment in M,
then a = F(a) is a curve segment in M with velocity a' = F*(a'). Since
F* preserves dot products, it preserves norms, so  a'  =  F*(a') J =
 F(a)'  =  a'  . Hence
L(a) = f  a' (t) \\dt= f  a'(t)  dt = L(a).
•'o J a
Now if a runs from p to q in M, its image a = F(a) runs from F(p)
to F(q) in M. Reciprocally, if is a curve segment in M from ^(p) to F(q)
in M, then F _1 (/8) runs from p to q in M. We have in fact established a
onetoone correspondence between the collection of curve segments used
to define p(p, q) and those used for p(F(p), F(q)). But as shown above,
corresponding curves have the same length, so it follows immediately
that p(p, q) = p(F(p), F(q)). I
Thus we may think of an isometry as bending a surface into a different
shape without changing the intrinsic distance between any of its points.
Consequently the inhabitants of the surface are not aware of any change at
all, for their geometric measurements all remain exactly the same.
If there exists an isometry from M to M, then these two surfaces are
said to be isometric. For example, if a piece of paper is bent into various
shapes without creasing or stretching, the resulting surfaces are all iso
metric (Fig. 6.10).
To study isometries it is convenient to separate the geometric condition
of preservation of dot products from the onetoone and onto requirements.
4.4 Definition A local isometry F: M — + N of surfaces is a mapping that
preserves dot products of tangent vectors (that is, F* does).
266
GEOMETRY OF SURFACES IN E 3
[Chap. VI
FIG. 6.10
Thus an isometry is a local isometry that is onetoone and onto.
If F is a local isometry, the earlier argument still shows that F is a regular
mapping. Then for each point p of M the inverse function theorem (5.4
of Chapter IV) asserts that there is a neighborhood It of p in M that F
carries diffeomorphically onto a neighborhood V of F(p) in N. Now "U
and V are themselves surfaces in E 3 , and thus the mapping F  It: 11 — > V
is an isometry. In this sense a local isometry is, indeed, locally an isometry.
There is a simple patch criterion for local isometries using the functions
E, F, and G defined in Section 4 of Chapter V.
4.5 Lemma Let F: M — * N be a mapping. For each patch x: D — > M,
consider the composite mapping
x = F(x):D^N.
Then F is a local isometry if and only if for each patch x we have
E = E
F = F
G = G
(Here x need not be a patch, but E, F, and G are defined for it as usual. )
Proof. Suppose the criterion holds — and only for enough patches to cover
all of M. Then by one of the equivalences in Exercise 1, to show that F*
preserves dot products we need only prove that
x u  = F*(x„)
x u 'x v = F*(x u )F*(x v ),
xll = \\F*M
But as we saw in Chapter IV, it follows immediately from the definition
of F* that ^*(x M ) = x„ and F*(x„) = x v . Thus the equations above fol
low from the hypotheses E = E, F = F, G = G. Hence F is a local iso
metry.
Reversing the argument, we deduce the converse assertion. 
This result can sometimes be used to construct local isometries. In the
simplest case, suppose that M is the image of a single patch x: D — > M .
Sec. 4] ISOMETRIES AND LOCAL ISOMETRIES 267
Then if y: D > N is a patch in another surface, define a mapping F:M*N
by
F(x(u,v)) = y(M,y) for (w,v) in D.
If
E = E,F = F,G = G,
then by the above criterion, F is a local isometry.
4.6 Example
(1) Local isometry of a plane onto a cylinder. The plane E 2 may be con
sidered as a surface, with natural frame field Ui, U*. If x: E — > M is a
parametrization of some surface, then Exercise 1 shows that x is a local
isometry if
x*(U t )'**(Ui) = UiUj,
Since x* (t/i) = x„, x* (C7 2 ) = x„, and U^Uj = «./, this is the same as re
quiring E = 1, F = 0, G = 1.
To take a concrete case, the parametrization
x(u, v) = (r cos f  J , r sin ( j , vj
of the cylinder M: x + y 2 = r 2 has E = 1, F = 0, G = 1. Thus x is a
local isometry which wraps the plane E 2 around the cylinder, with hori
zontal lines going around crosssectional circles, and vertical lines to rulings
of the cylinder.
(2) Local isometry of a helicoid onto a catenoid. Let H be the helicoid which
is the image of the patch
x(u,v) = (u cos v, u sin v, v).
Furnish the catenoid C with the canonical parametrization y: E 2 — ► C
discussed in Example 6.4 of Chapter V. Thus
y(u,v) = (g(u), h(u) cosy, h(u) sin v)
g(u) = sinh 1 w h(u) = \/l + u 2 .
Let F: H — » C be the mapping such that
F(x («,»)) = y(u,v).
To prove that F is a local isometry, it suffices to check that
E = 1 = E, F = = F, G = 1 + u 2 = h 2 = G.
F carries the rulings (v constant) of H onto meridians of the surface of
revolution C, and wraps the helices (u constant) of H around the parallels
268
GEOMETRY OF SURFACES IN E 3
[Chap. VI
of C. In particular, the central axis of H (z axis) is wrapped around the
minimal circle x = of C.
Figure 6.11 shows how a sample strip of H is carried over to C.
Suppose that the helicoid H (or at least a finite region of it) has been
stamped like an automobile fender out of a flexible sheet of steel— the
patch x does this. Then H may be wrapped into the shape of a catenoid
with no further distortion of the metal (Exercise 5 of Section 5).
A similar experiment may be performed by cutting a hole in a pingpong
ball representing a sphere in E 3 . Mild pressure will then deform the ball
into various nonround shapes, all of which are isometric. For arbitrary
isometric surfaces M and M in E 3 , however, it is generally not possible
to bend M (through a whole family of isometric surfaces) so as to produce
M.
There are special types of mappings other than (local) isometries that
are of interest in geometry.
4.7 Definition A mapping of surfaces F: M — > N is conformal provided
there exists a realvalued function X > on M such that
\\F*M\\ =X(p)v p 
for all tangent vectors to M. The function X is called the scale factor of F.
Note that if F is a conformal mapping for which X has constant value 1,
F is a local isometry. Thus a conformal mapping is a generalized isometry
FIG. 6.11
Sec. 4] ISOMETRIES AND LOCAL ISOMETRIES 269
for which lengths of tangent vectors need not be preserved — but at each
point p of M the tangent vectors at p all have their lengths stretched by
the same factor.
The criteria in Lemma 4.5 and in Exercise 1 may easily be adapted
from isometries to conformal mappings by introducing the scale factor
(or its square). In Lemma 4.5, for example, replaced = E by E\ 2 (x) = E,
and similarly for the other two equations.
An essential property of conformal mappings is discussed in Exercise 8.
EXERCISES
1. If F: M — > N is a mapping, show that the following conditions on its
derivative map at one point p are logically equivalent :
(a) F* preserves inner products.
(b) F* preserves lengths of tangent vectors, that is,  F* (v)  =  v 
for all v at p.
(c ) F* preserves frames : If ei, e^ is a tangent frame at p, then
F*M,F*M
is a tangent frame at F(p).
(d) For some one pair of linearly independent tangent vectors v and
w at p
II *■*(▼) II = II v , F*(w)    w , and F*(v).F*(w) = vw
[Hint: It suffices, for example, to prove that (a) ==> (c) => (d) ==» (b) => (a).]
These are general facts from linear algebra; in this context they provide
useful criteria for F to be a local isometry.
2. Show that each of the following conditions is necessary and sufficient
for F: M — > N to be a local isometry.
(a) F preserves the speeds of curves:  F(a)'  = \\ a  for all curves
a in M.
(b) F preserves lengths of curves: L(F(a)) = L(a) for all curve
segments am. M.
3. Let M be the xy plane in E 3 with the origin removed. Show that the
intrinsic distance from ( — 1,0,0) to (1,0,0) in M is 2, but that there
is no curve segment in M which joins these points and has length 2.
(Hint: Ex. 11 of II. 2.)
4. Formulate precisely and prove: Local isometries can shrink but not
increase intrinsic distance.
270 GEOMETRY OF SURFACES IN E 3 [Chap. VI
5. Let a,/3 : / ■—> E 3 be unitspeed curves with the same curvature function
k > 0, and assume that the ruled parametrization
x(u,v) = a(u) + vT(u)
of the tangent surface of a is actually a patch. Find a local isometry
from the tangent surface of a to :
(a) The tangent surface of 0.
(b) A region D in the plane.
(Hint: Ex. 9 of III. 5.)
6. Show that the preceding exercise applies to the tangent surface of a
helix and find the image region D in the plane.
7. Modify the conditions in Exercise 1 so that they provided criteria
for F to be a conformal mapping. Then prove that a patch x: D —* M
is a conformal mapping if and only if E = G and F = 0.
8. Show that a conformal mapping F: M — > N preserves angles in this
sense: If # is an angle (0 ^ # ^ x) between v and w at p, then # is
also an angle between F* (v) and F# (w) at F(p).
9. If F: M — » Af is an isometry, prove that the inverse mapping
i* 11 : M — > M is also an isometry. If F: M — > N and G: N + P are
(local) isometries, prove that the composite mapping GF: Af — » P is a
(local) isometry.
10. Let x be a parametrization of all of M, x a parametrization in N.
If F: M — > iV is a mapping such that F(x(u, v)) = x(/(w), fif(y)), then
(a) Describe the effect of F on the parameter curves of x.
(b) Show that F is a local isometry if and only if
(In the general case, / and gr are functions of both u and y, and this
criterion becomes more complicated.)
(c) Find analogous conditions for F to be a conformal mapping.
11. Let M be a surface of revolution, and let F: H — > M be a local isometry
of the helicoid which (as in Example 4.6) carries rulings to meridians,
and helices to parallels. Show that M must be a catenoid. (Hint: Use
Exercise 10.)
12. Let M be the image of a patch x with E = 1, F = 0, and G a function
of u only (G v = 0). If the derivative d(s/G)/du is bounded, show
that there is a local isometry of M into a surface of revolution.
Thus any small enough region in M is isometric to a region in a
surface of revolution.
Sec. 5] INTRINSIC GEOMETRY OF SURFACES IN E 3 271
13. Let x be the geographical patch in the sphere 2 of radius r (Example
2.2 of Chapter IV). Stretch x in the northsouth direction to produce
a conformal mapping. Explicitly, let
y(u,v) = x(u,g(v)) with 0(0) = 0,
and determine g so that y is a conformal. Find the scale factor of y
and the domain D such that y(D) omits only a semicircle of 2. (Mer
cator's map of the earth derives from y: its inverse is Mercator's pro
jection.)
14. Show that stereographic projection P: S > E 2 (Example 5.2 of Chapter
IV) is conformal, with scale factor
15. Let M be a surface of revolution whose profile curve is not closed,
hence has a onetoone parametrization. Find a conformal mapping
F: M — > E 2 such that meridians go to lines through the origin and
parallels go to circles centered at the origin.
5 Intrinsic Geometry of Surfaces in E 3
In Chapter III we defined Euclidean geometry to consist of those concepts
preserved by Euclidean isometries. The same definition applies to surfaces:
The intrinsic geometry of M C E 3 consists of those concepts — called iso
metric invariants — that are preserved by all isometries F: M — > M. For
example, Theorem 4.3 shows that intrinsic distance is an isometric in
variant. We can now state Gauss's question (mentioned in the start of
the chapter) more precisely: Which of the properties of a surface M in E
belong to its intrinsic geometry? The definition of isometry (Definition 4.2)
suggests that isometric invariants must depend only on the dot product
as applied to tangent vectors to M. But the shape operator derives from a
normal vector field, and the examples in Section 4 show that isometric
surfaces in E 3 can have quite different shapes. In fact, these examples
provide a formal proof that shape operators, principal directions, principal
curvatures, and mean curvature definitely do not belong to the intrinsic
geometry of M <Z E .
To build a systematic theory of intrinsic geometry, we must look back
at Section 1 and see how much of our work there is intrinsic to M . Using
the dot product only on tangent vectors to M , we can still define a tangent
frame field E u E 2 on M. Thus from an adapted frame field we can salvage
the two tangent vector fields E u E 2 — and hence also their dual 1 forms
272
GEOMETRY OF SURFACES IN E 3
[Chap. VI
0i, 2 . It is somewhat surprising to find that these completely determine
the connection form a>i 2 .
5.1 Lemma The connection form o) U = — co 2 i is the only 1form that
satisfies the first structural equations
dd\ = CO12 A 2 ddi = CO21 a 1#
Proof. Apply these equations to the tangent vector fields E h E 2 . Since
6i{Ej) = d ih the definition of wedge product (Definition 4.3 of Chapter
IV) gives
"12 (#1) = dd^Ex.Et)
(aa(E t ) = co2i(# 2 ) = dd 2 (E 1 ,E i ).
Thus by Lemma 2.1, a>i 2 = — «2i is uniquely determined by 0i, 2 . 
5.2 Remark In fact, this proof shows how to construct 0012 = — a> 2 i
without the use of Euclidean covariant derivatives (as in Section 1).
Given Ei, E2 and thus 0i, 2 ; take the equations in the above proof as the
definition of a>i 2 on E\ and E2. Then the usual linearity condition
"12(F) = coi 2 (yi£'i + V2E2) = Via)i2(Ei) + V20in(,E2)
makes u>n a 1form on M, and one can easily check (by reversing the argu
ment above) that coi 2 = — «2i satisfies the first structural equations.
If F: M — > M is an isometry, then we can transfer a tangent frame field
Ei, E 2 on M to a tangent frame field E u E 2 on M: For each point q in M
there is a unique point p in M such that F(p) = q. Then define
£i(q) = nWp))
^ 2 (q) = F*(E 2 (p)).
(Fig. 6.12).
FIG. 6.12
Sec. 5] INTRINSIC GEOMETRY OF SURFACES IN E 3 273
In practice we shall abbreviate these formulas somewhat carelessly to
Ei = F*{E\), E 2 = F*{E 2 ).
Because F* preserves dot products, Ei, E 2 is a frame field on M, since
EfEi = F*(Ei).F*(Ei) = EfEj = 5 iy .
5.3 Lemma Let F: M — » M be an isometry, and let Ei, E2 be a tangent
frame field on M. If E u E 2 is the transferred frame field on M, then
(1) e x = F*(e x ), e 2 = F*(e 2 )
(2) co 12 = F*(a 12 ).
Proof. (1) It suffices to prove that 0* and F*(di) have the same value on
Ei and ^2. But for 1 ^ i,j S 2 we have
F*(9i)(Ej) = &(F*0,) = 0~i(#;) = «« = BiiEi).
(2) Consider the structural equation ddi = W12 a 2 on M. If we apply
F*, then by the results in Chapter IV, Section 5, we get
d(F*6 1 ) = F*(ddi) = F*(«u) a F*(d 2 ).
Hence, by (1), we have
dOx = F*{u n ) a 2 .
The other structural equation
G?0 2 = C021 A dl
gives a corresponding equation, so
ddi = F*(w 12 ) a 2
d6 2 = F*(wa) a 0l
But now (2) is an immediate consequence of the uniqueness property
(Lemma 5.1), since
F*(«a) = F*(«w) = F*(«w). I
From this rather routine lemma we easily derive a proof of the celebrated
theorema egregium of Gauss.
5.4 Theorem Gaussian curvature is an isometric invariant; that is if F:
M —> M is an isometry, then
K(p) = K(F(p))
for every point pinM.
Proof. For an arbitrary point p of M, pick a tangent frame field E h E 2
274 GEOMETRY OF SURFACES IN E 3 [Chap. VI
on some neighborhood of p and transfer via F* to E if E 2 on M. By the
previous lemma, F*(wn) = co x2 . According to Corollary 2.3, we have
doiu = —Kdi a 02.
Apply F* to this equation. By the results in Chapter IV, Section 5, we get
d(F*«! 2 ) = F*(dun) = F*(K)F*(d 1 ) a F*(d 2 )
where F*(K) is simply the composite function K(F). Thus by the pre
ceding lemma,
dwi 2 = —K(F)6 1 a 2 .
Comparison with dw 12 = — Kdi a 2 yields K = K(F); hence, in particular,
K(p) = K(F(p)). 
Gauss's theorem is one of the great discoveries of nineteenthcentury
mathematics, and we shall see in the next chapter that its implications are
farreaching. The essential step in the proof is the second structural equa
tion
do)\2 = — Kdi a 6%.
Once we prove Lemma 5.1, all the ingredients of this equation, except K,
are known to derive from M alone — thus K must also. This means that
the inhabitants of M CI E can determine the Gaussian curvature of their
surface even though they cannot generally find S and have no conception
of the shape of M in E 3 .
The machinery of differential forms puts this heuristic reasoning beyond
doubt by supplying the formal proof of isometric invariance in Theorem
5.4. This remarkable situation is perhaps best illustrated by the formula
K = kik2,: An isometry need not preserve the principal curvatures, nor
their sum, but it must preserve their product. Thus the shapes which isometric
surfaces may have — although possibly quite different — are by no means
unrelated.
A local isometry is, as we have shown, an isometry on all sufficiently
small neighborhoods. Thus it follows from Theorem 5.4 that local isometries
preserve Gaussian curvature. For example, in Example 4.6 the plane and
the cylinder both have K = 0. (This is why we did not hesitate to call the
curved cylinder "flat". Intrinsically it is as flat as a plane.) In the second
part of Example 4.6, at corresponding points
x(u,v) and F(x(u,v)) = y(u,v),
the helicoid and catenoid have exactly the same Gaussian curvature:
1/(1 + uf (see Examples 4.3 and 6.4 of Chapter V.)
Sec. 5] INTRINSIC GEOMETRY OF SURFACES IN E 3 275
Gauss's theorema egregium can obviously be used to show that given
surfaces are not isometric. For example, there can be no isometry of the
sphere 2 (or even a very small region of it) onto part of the plane, since
their Gaussian curvatures are different. This is the dilemma of the map
maker: The intrinsic geometry of the earth's surface is misrepresented by
any flat map.
The next section is computational; Section 7 will provide more isometric
invariants.
EXERCISES
1. Geodesies belong to intrinsic geometry: If a is a geodesic in M, and
F: M — » N is a (local) isometry, then F(a) is a geodesic of N. (Hint: See
Ex. lof VI.l.)
2. Use Exercise 1 to derive the geodesies of the circular cylinder (Example
5.8 of Chapter V). Generalize to an arbitrary cylinder.
3. For a (connected) surface, the values of its Gaussian curvature fill an
interval. If there exists a local isometry of M onto N (in particular if
M and N are isometric), show that M and N have the same curvature
interval. Give an example to show that the converse is false.
4. Prove that no two of the following surfaces are isometric: sphere, torus,
helicoid, circular cylinder, saddle surface.
5. Bending of the helicoid into the catenoid (4.6). For each number t in the
interval ^ t ^ ir/2, let x t : E 2 — » E be the mapping such that
x t (u,v) = cos t(Sc, Ss, v) + sin t( — Cs, Cc, u)
where C = cosh u, S = sinh u, c = cos v, and s = sin v.
Now x is a patch covering the helicoid, and x x/2 is a parametrization
of the catenoid — these are mild variants of our usual parametrizations,
and the catenoid now has the z axis as its axis of rotation. If we imagine
t to be time, then x t for ^ t ^ x/2 describes a bending of the helicoid
Mo in space which carries it onto the catenoid M^/2 through a whole
family of intermediate surfaces M t = x t (E 2 ). Prove
(a) M t is a surface. (Show merely that x t is regular.)
(b) M t is isometric to the helicoid M if t < ir/2. (Show that F t : M — > M t
is an isometry, where
F t (x (u,v)) = x t (u,v).
Also show that for t — ir/2, F T ft is a local isometry. )
(c) Each M t is a minimal surface. (Compute x uu + x vv = 0.)
276 GEOMETRY OF SURFACES IN E 3 [Chap. VI
(d) Unit normals are parallel on orbits: Along the curve t — > x t {u, v) by
which the point x (u, v) of M moves to M T/2 , the unit normals U t
of successive surfaces are parallel.
(e) Gaussian curvature is constant on orbits. (Find K t (x t (u,v)), where
K t is the Gaussian curvature of M t ).
A brilliant series of illustrations of this bending is given in Struik [6].
6. Show that every local isometry of the helicoid H to the catenoid C must
carry the axis of H to the minimal circle of C, and the rulings of H to
the meridians of C, as in Example 4.6. (Compare Ex 11 of VI.4.)
6 Orthogonal Coordinates
We have seen that the intrinsic geometry of a surface M C E 3 may be
expressed in terms of the dual forms 6 ly 2 , and connection form o>i 2 derived
from a tangent frame field E u E 2 . These forms satisfy
the first structural equations:
dd\ = Wi2 A 6 2
dd 2 = a>2i A $i
the second structural equation:
dun = —Kd\ a 02
In this section we develop a practical way to compute these forms — and
hence a new way to find the Gaussian curvature of M.
The starting point is an orthogonal coordinate patch x: D — » M, one for
which F = x u 'x v = 0. Since x u and x„ are orthogonal, dividing by their
lengths  x„  = \/E and  x„  = y/G will produce frames.
6.1 Definition The associated frame field Ei, E% of an orthogonal patch
x: D — > M consists of the orthogonal unit vector fields E\ and E 2 whose
values at each point x(u, v) of x(D) are
x„(w, v)/\/E{u,v) and x v (u,v)/\/G(u,v).
In Exercise 9 of Section 4 of Chapter IV, we associated with each patch
x the coordinate functions u and v, which assign to each point x(u, v) the
numbers u and v, respectively. For example, for the geographical patch x
of Example 2.2 of Chapter IV, the coordinate functions are the longitude
and latitude functions on the sphere 2. In the extreme case when x is the
identity map of E 2 , the coordinate functions are just the natural coordinate
functions (u, v) — > u, (u,v) — > v on E .
Sec. 6] ORTHOGONAL COORDINATES 277
For an orthogonal patch x with associated frame field E\, E%, we shall
express 6 X , 2 , and coi 2 in terms of the coordinate functions u, v. Since x is
fixed throughout the discussion, we shall run the risk of omitting the in
verse mapping x 1 from the notation. With this convention, the coordinate
functions u = w(x 1 ) and y = f(x _1 ) are written simply u and v, and
similarly x u and x„ now become tangent vector fields on M itself. Thus the
associated frame field of x has the concise expression
Er = x u /\/E E 2 = x v /VG. (1)
Now the dual forms 0i, 2 are characterized by di(Ej) = da, and in the
exercise referred to above it is shown that
du(x u ) = 1 dv(x u ) =
du(x v ) = dv(x v ) = 1.
Thus we deduce from (1) that
0i = y/Edu d 2 = y/Gdo. (2)
By using the structural equations we shall find analogous formulas for
W12 and K. Recall that for a function /, df = f u du + f v dv, where the sub
scripts indicate partial derivatives. Hence from (2) we get
dB x = d(VE) a du = (VE)v dv du = ~~(vE) v ^ a ^
VG
d6 2 = d(\/G) a dv = (VG)u du dv = ~WG)u dv A ^
\/E
where we have used the alternation rule for wedge products and substituted
dv = 2 /"\/(r and du = di/y/E from (2 ) . Comparison with the first struc
tural equations dd\ = wi 2 a 2 and dd 2 = — coi 2 a di shows that
m2 = ^h^ du + (VG), dv (3 )
Vg Ve
The logic is simple: By the computations above, this form satisfies the
first structural equations; hence by uniqueness (Lemma 5.1), it must be W12.
6.2 Example Geographical coordinates on the sphere. For the geographi
cal patch x in the sphere 2 (Example 2.2 of Chapter IV), we have com
puted E = r 2 cos 2 v,F = 0,G = r 2 . Thus by formula (2) above,
di = r cos v du 2 = r dv.
Now (\/E) v = — r sin v and {\/G)u = 0; hence by (3),
wi 2 = sin v du.
278 GEOMETRY OF SURFACES IN E 3 [Chap. VI
The associated frame field of this patch is the same one we got in Ex
ample 1.6 from the spherical frame field in E 3 . With the notational shift
u —> &, v — > <p, the forms above are (necessarily) also the same. But now
we have a simple way to compute them directly in terms of the surface
with no appeal to the geometry of E 3 .
Finally we derive a new expression for the Gaussian curvature. In this
context, exterior differentiation of o>i 2 as given in (3 ) yields
du n = — ((\/E) v /\^G)v dv du + ((\/G)u/\/E) u du dv.
From (2) we get
0i a 2 = y/EG du dv;
hence
1
— dvdu = du dv = ~T^p 0i A 02
Thus the formula above becomes
dd)12 =
vm {(w"). + (w)J * A 6l
We now compare this with the second structural equation,
cfai2 = — K6x a 2 .
6.3 Lemma If x: D — » M is an orthogonal patch, then the Gaussian
curvature K is given in terms of x by
K =
veg \\ y/E )u\ vg a;
By contrast with the formula for K in Corollary 4.1 of Chapter V, the
functions £, m , and « (which describe the shape operator) no longer appear.
Indeed since K is now expressed solely in terms of E, F, and G, we have,
using Lemma 4.5, another proof of the isometric invariance of Gaussian
curvature.
In applications it is generally easier to repeat the derivation of Lemma
6.3 in each case, rather than look it up and substitute in it. For example,
consider the polar parametrization x(u,v) = (u cos v, u sin v ) of the Euclid
ean plane E 2 . Here E = 1, F = (so x is orthogonal), and G = u. Thus
by (2), 0i = du and 2 = u dv. Since
Sec. 6]
ORTHOGONAL COORDINATES
279
ddi = and ddi = du dv = —dv a lt
we find cow = dv. But then do>i 2 = 0, which shows again that E 2 is flat.
6.4 Example The natural frame field of a surface of revolution. For
a canonical parametrization
x(u, v) = (g{u), h(u) cos v, h(u) sin v)
of a surface of revolution, the associated frame field has E x in the direction
of the meridians and E* in the direction of the parallels (Fig. 6.13). Since
x is orthogonal, with E = 1, G = h 2 , we get 0i = du and 2 = h dv. Here
/i is a function of w alone, so h v = 0, and h u is the ordinary derivative h .
From (3) — or by direct computation — o>i 2 = h dv, so
h"
dwi2 = h" du dv = — 0i a 2
We conclude that the Gaussian curvature is K = h"/h, in agreement with
the results of Lemma 6.3 of Chapter V.
EXERCISES
1. Compute the dual forms, connection form, and Gaussian curvature for
the associated frame field of the following orthogonal patches:
(a) \(u, v) = (u cos v, u sin v, bv), helicoid
(b) x(u,v) = (u cos v, u sin v, u 2 /2), paraboloid of revolution.
(c) x(u,v) = (u cos v, u sin v, au), cone
2. Let the patch x: D — » M be a conformal mapping. (The associated
coordinate system is said to be isothermal.) Prove:
(a) K = —A (log E)/2E, where A is the Laplacian: A/ = f uu + /„ .
FIG. 6.13
280
GEOMETRY OF SURFACES IN E 3
En
[Chap. VI
Ex =
FIG. 6.14
(b) The mean curvature H is zero if and only if x uu + x„„ = 0. (Hint
Use Ex. 7 of VI.4.)
3. For a patch with E = G = 1, show that K = — ^ uv /sm #, where & is
the coordinate angle. (Hint: For the frame field with Ex = x u as in
Fig. 6.14, show that X = du + cos $ dv, 2 = sin & dv.)
A. If x is a principal patch (Ex. 9 of V.4), prove
(a)
0013 =
Ve
du
V'G
(b) u = HE V
= HG U .
5. By refining the argument in the text, show that equations (2) and (3)
are literally true provided d u 2 , and con are replaced by their coordi
nate expressions
x*(0i), x*(0 2 ), x*(coi 2 ).
7 Integration and Orientation
The main aim of this section is to define the integral of a 2form over a
compact oriented surface. This notion does not involve geometry at all;
it belongs to the integral calculus on surfaces (Chapter IV, Section 6).
However, we shall motivate the definition by considering some related
geometric problems.
Perhaps the simplest use of double integration in geometry is in finding
the area of a surface. To discover a proper definition of area, we start with
a patch x: D — > M and ask what the area of its image x(D) should be. Let
AR be a small coordinate rectangle in D with sides Am and Av. Now x
distorts AR into a small curved region x(AR) in M, marked off in an obvious
way by four segments of parameter curves, as shown in Fig. 6.15.
Sec. 7]
INTEGRATION AND ORIENTATION
281
(AR)
FIG. 6.15
We have seen that the segment from x (w, ») to x(m + Au, v) is linearly
approximated by Au x u (evaluated at (u, v)), and the one from x(u, v) to
x(u, v + Av) by Ay x„. Thus the region x(A.R) is approximated by the
parallelogram in T X ( U<v) (M) with these vectors as sides. From Chapter II,
Section 1, we know that this parallelogram has area
 Aux u X Avx v  =  x u X x„  AuAv = s/EG — F 2 AuAv
We conclude that the area of x(AR) should be approximately s/EG — F 2
times the area AuAv of AR. At each point (u, v) the familiar expression
■\/EG — F 2 gives the rate at which x is expanding area at (u, v). Thus it is
natural to define the area of the whole region x(D) to be
fj VEG  F 2 du dv
Such integrals, of course, may well be improper; we shall avoid this
difficulty by modifying the notion of patch.
7.1 Definition The interior R° of a rectangle R: a ^ u ^ b, c ^ v S d
is the open set a < u < b, c < v < d. A 2segment x: R — » M is patchlike
provided the restricted mapping x: R° — * M is a patch in M.
The remark preceding Lemma 7.3 of Chapter IV shows that the area of
x(R) is finite, since it implies that \/EG — F 2 ^ is bounded on R.
A patchlike 2segment x: R — » M need not be onetoone on the bound
ary of R, so its image may not be very rectangular. In fact we shall now
see that the area of an entire compact surface may often be computed by
covering it with a single 2segment.
7.2 Example Areas of surfaces
(1) The sphere S of radius r. If the formula defining the geographical
patch is applied to the rectangle R: —t < u < w, — tt/2 < v < ir/2, we
obtain a 2segment that covers the whole sphere. Now
E = r 2 cos 2 v, F = 0, and G = r\
282
GEOMETRY OF SURFACES IN E 3
[Chap. VI
SO
VEG — F 2 = r 2 cos v,
and the area of the sphere is thus
• IT t/2
/ r 2 cos v du dv = 47ir 2
it Jt/2
(2) Torus 7 1 of radii R > r > 0. From Example 2.6 of Chapter IV we
can derive a patchlike 2segment covering the torus. Here
VEG  F 2 = r(R + r cos u),
so the area is
A{T) = I f r(R + r cos u) du dv = ItRt
(3) The bugle surface (Example 6.6 of Chapter V). Every surface of
revolution M has a canonical parametrization with E = 1, F = 0, G = h .
On a rectangle jR:a^w^6,0^w^ 2ir, x is a patchlike 2segment
whose image is the region of M between the parallels u = a and u = b
(Fig. 6.16). Thus the area of this region is
A ah = I I h du dv = 2t I h du.
J a •'0 J a
For the bugle surface, we saw in Chapter V that h(u) = ce~ ufc ; hence
A a b = 2irc I e u,c du
•'o
c\ 2/ —ale — b/c\
FIG. 6.16
Sec. 7] INTEGRATION AND ORIENTATION 283
FIG. 6.17
To find the area of the whole bugle — a noncompact surface — we expand
this region, letting a — > and b — » « .
Thus (see Remark 7.6) the bugle has finite area
lim A a b = 2irc
b»oo
To define the area of a complicated region we shall not try to fit a single
patchlike 2segment onto it. Instead we follow the usual scheme of ele
mentary calculus and break the region into simple pieces, then add their
areas.
7.3 Definition A paving of a region (P in M consists of a finite number of
patchlike 2segments xi, • • • , x k whose images fill (P in such a way that
each point of (P is in at most one set x t  (Ri° ) .
In short, the images of the x/s cover <P exactly and overlap only on their
boundaries (Fig. 6.17).
Not every region is pavable; since pavings are finite, compactness
is certainly necessary (Definition 7.2 of Chapter IV). It is safe to assume
that a compact region is pavable if its boundary consists of a finite number
of regular curve segments. In particular, an entire compact surface is always
pavable. t The area of a pavable region (P is defined to be the sum of the areas
of xiCRi), • • • , Hk(Rk) for a paving of (P. (The consistency problem here
will be discussed following the analogous definition, 7.5.)
The preceding exposition shows that computation of area does not de
mand differential forms, but integration of 2forms (Definition 6.3 of
Chapter IV) will give area — and much more besides. The first question is
this: Which 2form should we integrate over a patchlike 2segment x to get the
area of its image? By definition,
// M = // n(x u , x„) du dv.
t See remarks and reference following Theorem 8.5 of Chapter VII.
284 GEOMETRY OF SURFACES IN E 3 [Chap. VI
Thus we want a 2form whose value on x„,x„ is
 x tt X x,  = VEG  F*.
In general, a 2form n such that
m(v, w) = ±  v X w  for all v, w
is called an area form. Such a form assigns to every pair of tangent vectors
v, w either plus or minus the area of the parallelogram with sides v and w.
7.4 Lemma A surface M has an area form if and only if it is orientable.
On a (connected) orientable surface M there are exactly two area forms,
which are negatives of each other. (We denote them by dM and —dM.)
Proof. If v and w are linearly independent, then  v X w  > 0; thus
area forms are nonvanishing. Hence, by Definition 7.4 of Chapter IV, a
nonorientable surface cannot have an area form.
Now suppose that M is an orientable surface in E 3 . The proof of Theorem
7.5 of Chapter IV actually establishes a onetoone correspondence be
tween normal vector fields on M and 2forms on M. If U is a unit normal,
then the associated 2form dM is an area form, since
dM(v, w) = U(p)»y X w = ±  v X w .
(In Fig. 6.18 this number is positive, since v X w points in the same direc
tion as U(p), but if v and w were reversed, we would get the negative sign.)
Thus the two unit normal vector fields on M determine the two area
forms dM and — dM on M. 
To orient an orientable surface is to pick one of its two area forms, since
that amounts to the same thing as picking one of its unit normals.
V X W
FIG. 6.18
Sec. 7] INTEGRATION AND ORIENTATION 285
Finding area is not a really typical integration problem, since area is
always positive. Thus to find area by integrating an area form we must be
careful about signs. Suppose x is a patchlike 2segment in a surface oriented
by an area form dM. By definition,
/ / dM = // dM(x u , x„) du dv.
Now there are two cases:
(1) If dM(x u , x„) > 0, we say that x is positively oriented. Then by defi
nition of area form,
dM(x u , x„) =  x u X x„ ;
hence // x dM is the area of x(R).
(2) If dM(xu, x v ) < 0, we say that x is negatively oriented. Then
dM(x u , x v ) = —  x u X x„ ,
hence J/ x dM is minus the area of x(R).
Thus to find the area of a pavable oriented region by integrating its
selected area form, we cannot use an arbitrary paving; the paving must be
positively oriented, that is, consist only of positively oriented patchlike 2
segments. Then
A(<?) = T,M*(Ri)) = Z//
dM.
Now we replace the area form by an arbitrary 2form to get the definition
we are looking for.
7.5 Definition Let v be a 2form defined on a pavable oriented region (P
in a surface. The integral of v over (P is
where xi, • • • , x k is a positively oriented paving of (P.
There is a consistency problem in this definition: We must know that
two different positively oriented pavings of (P give the same value for the
sum on the right. A detailed proof would be somewhat long; the general
scheme is given on page 103 of Hicks [5].
Since compact surfaces are pavable, the definition above gives in particu
lar the integral of a 2form over a compact oriented surface.
7.6 Remark Improper integrals. We have defined area and the integra
tion of forms for compact surfaces; however, the notion of area can easily
286 GEOMETRY OF SURFACES IN E 3 [Chap. VI
be extended to a noncompact surface N. We define the area of N to be the
least upper bound of the set of all areas of payable regions (P in N:
A(N) = l.u.b. A(<P).
Thus A(N) = +00 if no finite upper bound exists.
By contrast, it is generally impossible to assign a value — finite or in
finite — to the improper integral /JV / dN. The special case / ^ 0, however,
may be handled in the same way as area; we set
ff fdN = l.u.b. If fdN ((P pavable in N)
JJ N JJ(p
For / ^ 0, switch to greatest lower bound. Thus values + oo and — °o are
possible in these two cases. Now let 6*1, (P 2 , • • • be a sequence of pavable
regions in N such that (P* is contained in (P f+1 , and every pavable region in
N is contained in some G\. It then follows that
lim ff fdN = ff f dN.
(We used the corresponding fact for area in (3) of Example 7.2.)
If (P is a pavable region in a surface M oriented by dM, we have seen
that ff<p dM is the area of (P. More generally, ffy f dM gives the integral
of a function / over (P — an obvious analogue of the usual integral J 6 / dx
from elementary calculus. We turn now to an important geometric applica
tion of this idea.
7.7 Definition Let K be the Gaussian curvature of a surface M, and let
(P be a pavable region in M oriented by dM. Then the number
IL
KdM
is called the total Gaussian curvature of (P.
When (P is an entire compact oriented surface M, we get the total Gaus
sian curvature of M. The total is an algebraic one: Negative curvature at
one place may cancel positive curvature at another.
To compute the total curvature, Definition 7.5 shows that it suffices
to know how to integrate the 2form K dM over patchlike 2segments. But
jf KdM = ff x*(K dM)
= ff K(x)x*(dM) = [ f K{xWEG  F 2 du dv
with the usual notation for x: R > M. Then K(x) may be computed
Sec. 7] INTEGRATION AND ORIENTATION 287
explicitly by Corollary 4.1 of Chapter V or by Lemma 6.3. Luckily the ori
entation problems here solve themselves automatically; see Exercise 4(c).
7.8 Example Total curvature of some surfaces
(1) Consfanf curvature. If the Gaussian curvature of M is constant, then
its total curvature is
ff KdM = K jf dM = K A(M).
Thus a sphere of radius r has total curvature 4ir (since K A(M) becomes
(l/r 2 )(4irr 2 )), and the bugle surface has total curvature — 2t (since
K A (M) becomes (  1/c 2 ) (2ttc 2 ) ) .
(2) Torus. Let x be the 2segment used on the torus T in Example 7.2.
By this example the area form dT has the coordinate expression
x*(dT) = y/EG  F 2 du dv = r(R + r cos u) du dv
But in Example 6.1 of Chapter V we computed
K(x) = (cos u)/r(R + r cos u)
for this same x. Hence, the torus has total curvature
J J KdT = j j cos u du dv = 0.
Thus on the torus the negative curvature of its inner half exactly balances
the positive curvature on its outer half, giving total curvature zero.
(3) Catenoid. This surface is not compact, and its area is infinite; never
theless its total curvature^ — treated by the remark above as an improper
integral — is finite. On the rectangle R: — a ^ u ^ a, O^vfS 2ir, the
parametrization in Example 6.1 of Chapter V becomes a patchlike 2
segment covering the region between the parallels u = —a and u = a.
(Fig. 6.19). From Example 6.1 of Chapter V, we get
K(x) = _1
c 2 cosh 4 (u/c)
and
x*(dM) — y/EGdudv = c cosh 2 (u/c).
Hence the region has total curvature
ff K JM — — f f^ ^ U ^ V — —4 t h (^\
JJ X J Jo c cosh 2 (u/c) \c /
288
GEOMETRY OF SURFACES IN E 3
[Chap. VI
FIG. 6.19
As a — * oo, this region expands to fill the whole surface; thus the total
curvature of the catenoid is
I
K dM = — 4x lim tanh
©
= — 4t.
The total curvatures computed above are 4x, — 2ir, 0, and — 4* — a
rather special set of numbers. Furthermore, none depends on the particu
lar "size" (radius r, constant c, • • • ) of its surface. A partial explanation is
provided by Corollary 7.10; a rather deeper one comes in Chapter VII,
Section 8.
If F: M — * iV is a mapping of oriented surfaces, then the Jacobian J of F
is the realvalued function on M such that
F*(dN) = J dM.
We can get an idea of the geometric meaning of J by arguing as in the
special case at the beginning of this section. If v and w are very small
tangent vectors at a point p of M, they span a parallelopiped in T P (M)
which approximates a small region AM in M . The character of the deriva
tive map F# is such that F*(v) and F*(w) are the sides of a parallelogram
in T F ( P) (N) approximating the image region F(AM), as shown in Fig.
6.20. By the definition of Jacobian we have
J(p)dM(y,w) = (F*dN)(v,w) = dN(F*v,F*w).
(*)
Now  v X w  is approximately the area of AM (and similarly for
F{AM)). Hence by taking absolute values we get
 J(p)  (area of AM) ~ area of F{AM).
Thus  J (p )  gives the rate at which F is expanding area at p. Furthermore
if AM is positively oriented, that is, dM(\, w) > 0, then (*) shows that
the sign of dN(F*\, F*w) is the same as that of J(p). Thus the sign of
Sec. 7] INTEGRATION AND ORIENTATION 289
**(w)
FIG. 6.20
«/(p) tells whether F preserves or reverses the orientation of AM.
In this context we call the number
ff JdM = If F*(dN)
the algebraic area of F(M). The discussion above shows that, roughly
speaking, each small region AM in M contributes to this total the algebraic
area of its image F(AM) :
(1) Positive, if the orientation of F(AM) agrees with that of N;
(2) Negative, if these orientations disagree (so F has turned AM over);
(3 ) Zero, if F collapses AM to a curve or a point.
Let us consider what this means in the case of the Gauss mapping (Exer
cise 4 of Section 1 of Chapter V).
7.9 Theorem The Gaussian curvature K of an oriented surface M C E 3
is the Jacobian of its Gauss mapping G: M — *■ 2.
(Here 2 is the unit sphere, oriented by the outward normal U or the
corresponding area form d2.)
Proof. If U = 2 gJJi is the unit normal orienting M, then the cor
responding Gauss mapping is G = (g it g 2 , ffr). Notice that if S is the shape
operator of M given by U, then
S(v) = V„£7 = 2vy[/ i (p)
and by Theorem 7.5 of Chapter I,
G*(v) =2Y[g i ]U i (G(p)).
Hence (r*(v) and — S(v) are parallel for any tangent vector v to M, as
shown in Fig. 6.21.
To prove the theorem we must show that
KdM = G*(d2),
so we evaluate these 2forms on an arbitrary pair of tangent vectors to M.
Using Lemma 3.4 of Chapter V,
290
GEOMETRY OF SURFACES IN E 3
[Chap. VI
U(p)
U(G(p))
G(P)
FIG. 6.21
(KdM)(v,w) = K(p) dM(v,w) = K(p) f/(p).v X w
= U(p)S(v) X £(w).
On the other hand,
(G*d2)(v,w) = d2(G*v, G*w) = U(G(p))G*v X G*w
Now a triple scalar product depends only on the Euclidean coordinates of
its vectors, so G*(\) and Cr*(w) may be replaced by the parallel vectors
— S(v) and — S(w). Furthermore, by the definition of G and the special
character of the unit sphere 2, the vectors C/(p) and U(G(p)) are also
parallel (Fig. 6.21). Thus the two triple scalar products above are the
same — and the proof is complete. 
7.10 Corollary The total Gaussian curvature of an oriented surface
M C E 3 is the algebraic area of the image G(M) of its Gauss mapping
G:M^2.
To prove this it suffices to integrate the form
KdM = G*(d2)
over M .
Algebraic area can be tricky when the mapping F: M —> N folds M
many times over the same regions in N. Thus for practical purposes, the
following special case of Theorem 7.10 is simpler, since it involves only
ordinary area.
7.11 Corollary If (R is an oriented region in M CZ E on which (1) the
Gauss map G is onetoone (U is not parallel at different points of (R), and
(2) either K ^ or K ^ 0, then the total curvature of (R is plus or minus
the area of G(6i), where the sign is that of K. Furthermore this area does
not exceed 47r.
(The proof uses improper integrals.) For example, consider the Gauss
mapping of an oriented torus. Now G collapses the top and bottom circles
of T (where K = 0) to the north and south poles of 2. If, as in Fig. 5.21,
and S are the outer and inner halves of T, then G maps (where K ^ 0)
Sec. 7]
INTEGRATION AND ORIENTATION
291
in onetoone fashion onto the whole sphere Z— and does the same for S
(where K ^ 0). Thus T has total curvature + A (2)  A (2) = 0, as we
found in Example 7.8 by an explicit integration.
As another example we find that the entire bugle surface B satisfies the
hypotheses in Corollary 7.11. In fact, as Fig. 6.22 suggests, the Gauss
mapping carries its profile curve in onetoone fashion onto a quarter of a
great circle in 2. Thus by moving U around the parallels of B, we see that
G is onetoone from B onto an open hemisphere (minus its central point,
since the rim is not part of B). Thus the total curvature of the bugle is
— () A (2) = — 2?r. Furthermore, since the bugle has constant curvature,
we can find its area without explicit integration: Total curvature — 2?r
divided by (constant) curvature K =  1/c gives the area 2ttc 2 , as found
in Example 7.2.
On an oriented surface, the ambiguity in the measurement of angles
mentioned in Chapter II, Section 1, can be reduced. If the surface is ori
ented by an unit normal U, then for every tangent vector v to M , U X v
is a tangent vector orthogonal to v. We shall think of U X v as v rotated
through +90°. Then if v and w are unit tangent vectors at a point of M,
a number # is defined to be an oriented angle from v to w provided
w = cos # v + sin & (U X v).
(Fig. 6.23). All oriented angles from v to w thus have the form & + 2im,
where n is an arbitrary integer. (The same scheme applies to any pair of
nonzero tangent vectors: It suffices to divide by their norms to get unit
vectors. )
Consistency is the essence of orientability— in studying an oriented
surface oriented by an area form dM we shall always use positively oriented
patches, dM(x u , x v ) > 0, and positively oriented frame fields, for which
dM(E u E 2 ) = +1.
Profile curve
FIG. 6.22
292
GEOMETRY OF SURFACES IN E 3
[Chap. VI
FIG. 6.23
(Note that by definition of area form the only possible values of dM on a
frame are ±1.) For a positively oriented frame field, we can now give
geometric meaning to the wedge product of its dual forms :
dM = 0! a 2
on the domain of the frame field. To prove this useful fact, it suffices to
note that both sides have the same value, +1, on E x , E 2 . The second struc
tural equation (Corollary 2.3) then becomes da>u = — K dM.
1. For a Monge patch
EXERCISES
x(u,v) = (u,v,f(u,v)),
show that the area of x(D) is given by the usual formula from ele
mentary calculus. Deduce that A(x(D)) ^ A(D).
2. Find a formula for the area of an arbitrary surface of revolution, and
interpret it as A = 2irLh, where L is the length of the profile curve and
h is its average distance from the axis of revolution (Pappus).
3. Find the area of the following surfaces:
(a) The region in the saddle surface z = xy covering the disc x 2 + y 2
^ c in the xy plane.
(b) Catenoid (c) The Mobius band (Ex. 7 of IV.7).
4. Let M be a compact surface oriented by dM; let — M be the same
surface oriented by the other area form —dM. Prove
(a) / / (civi + C2P2) = c\ I J vi + c 2 / J V2 (ci, c 2 constant).
Sec. 7] INTEGRATION AND ORIENTATION 293
<»/L'~/JL
(Hint: If x(u, v) = x(v, u), show that x and x have opposite orienta
tions; then look ahead to Exercise 21.)
(c) IL fdM = IL M f( ~ dM) 
(d) UJ^^ythenff fdM£ fj g dM
(Note the effect of / = or g = 0.)
Property (c) shows how to define the integral of a, function over a
compact surface that is merely orienta&Ze; either choice of area form
leads to the same result. In particular, the total curvature of a com
pact orientable surface is now welldefined.
Total curvature of surfaces of revolution. On a surface of revolution M
with profile curve a, let Z a & be the region ("zone") between the paral
lels through a (a) and a (b).
(a) Show that the total curvature of Za& is 27r(sin <p a — sin <pb), where
these are the slope angles of a at a (a) and a (b) — measured rela
tive to the axis of revolution (Fig. 6.24).
(b) Deduce that every surface of revolution whose profile curve is
closed has total curvature zero.
If the profile curve a is not closed, then it is onetoone on some open
interval A < t < B. Define the total curvature in this case to be
27r(lim a A sin <p a — linuB sin v?&)
provided both limits exist.
(c) Test this formula on the bugle and catenoid.
a'ib)
FIG. 6.24
294 GEOMETRY OF SURFACES IN E 3 [Chap. VI
6. Show that the Gauss mapping of a surface M (Z E is conformal if
and only if M is part of either a sphere or a minimal surface.
7. Find the total curvatures of the constant curvature surfaces of revolu
tion (Chapter V, Section 6, and Exercises); deduce their areas.
8. The area forms of E 2 are, as expected, ± du dv (since du and dv are
the dual forms of a frame field). The natural orientation of E 2 is by du dv;
this orientation is assumed unless the contrary is explicitly stated.
(a) Using the general definition in the text, show that the Jacobian
of a mapping
F = (/,<?): E 2 *E 2
is given by the usual formula
" = JuQv JvQu
(b) Show that the Jacobian of a patch x: D — » M in an oriented sur
face (D connected) is ± y/EG — F 2 , where the sign depends on
whether x is positively or negatively oriented.
9. Let M be a ruled surface whose rulings are entire straight lines, and
assume that K < 0.
(a) Show that the total curvature of M is —2L(8), where 5 is the
director curve, with  5  = 1.
(b) Compute the total curvature of the saddle surface M : z = xy
by this method, and check the result using Corollary 7.11.
10. Total curvature of quadric surfaces
(a) Find the total curvature of an arbitrary hyperbolic paraboloid,
elliptic paraboloid, and ellipsoid.
(b) Show that the total curvature of the hyperboloid of revolution
M: (x 2 + y)/a  z/c = 1 is lira/Va 2 + c 2 .
Thus the total curvature of an elliptic hyperboloid is dependent on
the particular "dimensions" of the surface — and the same is true for
an elliptic hyperboloid of two sheets.
11. A simple region S in M is a region that can be expressed as the image
F(D) of the disc u 2 + v ^ 1 in E 2 under a onetoone regular mapping
F. Show that S can be paved by a single patchlike 2segment x in
such a way that for any 1form we have
//.*/.
<t>,
where a is an edge curve of x. (Hint: See Ex. 12 of IV.6.)
12. (Continuation)
Sec. 7] INTEGRATION AND ORIENTATION 295
(a) If S is a simple region in E 2 , show that the area of S is
 / (u dv — v du) ,
where a is the "boundary curve" of S.
(b) Find the area of the region in E 2 enclosed by the ellipse
2 / 2 I 2 /7 2 i
u /a + v /b = 1.
13. Exercise 7 of Section 8, Chapter VII, will show that if <£ is any 1form
on a compact oriented surface, then
I
= 0.
Combine this result with Exercise 3 of Section 2 to show that if h
is the support function of M C E , then
A(M) + ff hH dM = ff HdM + // hK dM = 0.
Check these formulas on a sphere of radius r, oriented by the outward
unit normal.
14. Write & = 4 (u, v) if # is an oriented angle from u to v. Show that
(a) If t? = 4 (u, v) and <p = 2$. (v, w), then & + <p = 2$. (u, w).
(b) If # = 2$.(u,v), then # = 2$_(v,u).
15. Let a: I — » M be a curve in an oriented surface M. If V and W are
nonzero tangent vector fields on a, show there is a differentiate
function & on / such that &(t) is an oriented angle from V(t) to JT (<)
for each tin I. We call # an angle function from V to W. Note that any
two differ by an integer multiple of 2x. (Hint: Reduce to Ex. 12 of
ILL)
16. A mapping F: M — > N is areapreserving provided the area of any
pavable region (R in M is the same as the area of its image F((R) in N.
(Note that such a mapping must be onetoone.) Show that:
(a) A onetoone mapping F: M — > N is areapreserving if
EG  F 2 = EG  F 2
for every patch x in M, with x = ^(x) in N. (Hint: Show that
JP carries pavings to pavings.)
(b) Isometries are areapreserving; isometric surfaces have the same
area. Include the noncompact case.
(c) The mapping (1) in Example 5.2 of Chapter IV is areapreserving
but not an isometry. Deduce the standard formula for the area
296 GEOMETRY OF SURFACES IN E 3 [Chap. VI
of a zone in the sphere. (It suffices in (a) to consider a single
parametrization x if it covers all of M.)
17. If F: M — > N is a mapping of oriented surfaces with Jacobian J, show
that:
(a) F is regular if and only if J is never zero.
(b) F is areapreserving if F is onetoone and J = ±1. (The con
verse is true also.)
(c) If F is an isometry, then J = ±1, but the converse is false.
Since we are dealing only with connected surfaces, (a) shows that all
such mappings F separate into two classes: orientationpreserving if J > 0,
orientationreversing if J < 0. Except for patches (Exercise 8), we use this
notion mainly in the easy case where F is an isometry.
18. If F: M — ► M is an isometry of oriented surfaces, show that F is orienta
tionpreserving if and only if any one of the following conditions hold:
(a) F*(dM) = dM.
(b) F*(U(p) X v) = U(F(p)) XF*(v) for all tangent vectors v to M
at p (U and U the unit normals orienting M and M ).
(c) C/(p).vXw = l/(F(p)).F*(v) Xf*(w) for all pairs of tangent
vectors.
(d ) For any positively oriented frame field E h E 2 on M , F* (E x ) , F* (E 2 )
is a positively oriented frame field on M .
19. If F: M — > N is an orientationpreserving diffeomorphism of compact
oriented surfaces, show that
w IL f ' m  //, *
for any 2form on N. (Hint: Ex. 8 of IV.6.)
(b) Deduce that total curvature is an isometric invariant for compact
orientable surfaces.
(c) Extend (b) to the noncompact case, assuming either K ^ or
K ^ 0.
20. Gauss mapping G of some minimal surfaces. Prove :
(a) Catenoid. G is onetoone and its image covers all the sphere
except two points.
(b) Helicoid. The image of G omits exactly two points of the sphere,
and each point of the image is hit by an infinite number of points
of the helicoid.
(c) Scherk's surface (Ex. 21 of V.4). Same as the helicoid except that
exactly four points are omitted. {Hint: Consider Z = Vg on one
of the vertical lines.)
What are the total curvatures of these surfaces?
Sec. 8] CONGRUENCE OF SURFACES 297
21. In an oriented surface M let x and y be patchlike 2segments with
the same image x(D) = y(E). For any 2form v show that
///  * II;
with plus sign if x and y have the same orientation (positive or negative),
minus sign if opposite orientation. (Hint : examine the sign of the Jacobian
of y _1 x and use the change of variables formula for double integrals.)
8 Congruence of Surfaces
Two surfaces M and M are congruent provided there is an isometry F
of E 3 that carries M exactly onto M. Thus congruent surfaces have the
same shape — only their positions in E 3 can be different. For example, any
two spheres of the same radius are congruent (use the translation carrying
one center to the other), and the surfaces
M: z = xy and M: z =
x + y
are congruent under a 45° rotation about the z axis.
To simplify the exposition, we shall assume that the surfaces we deal
with in this section are orientable as well as connected.
8.1 Theorem If F is a Euclidean isometry such that F(M) = M, then
the restriction of F to M is an isometry F = F  Af: M — ► M of surfaces.
Furthermore, if M and M are suitably oriented, then F preserves shape
operators; that is,
F*{S(j)) = S(F*(v))
for all tangent vectors v to M.
In short, congruent surfaces are isometric and have essentially the same
shape operators. We emphasize, however, that isometric surfaces need not
be congruent, since, as we have seen, they may have quite different shapes
E3
Proof. We know from Chapter IV, Section 5, that the restriction
F: M — > M is a mapping. Furthermore the derivative maps of F and F agree
on tangent vectors to M. In fact, if v is tangent to M , then v is the initial
velocity of some curve a. in M — and since F = F  M , we have
F(a) = F(a).
Thus
F*(v) = F(a)'(0) = F(a)'(0) = F*(v).
298
GEOMETRY OF SURFACES IN E 3
[Chap. VI
It follows immediately that F* preserves dot products of tangent vectors
to M, for F* has this property for all pairs of tangent vectors (Corollary
2.2, Chapter III). Also, F: M —*■ M is onetoone (since F is) and onto
(by hypothesis); hence F is an isometry of surfaces.
Finally, we show that F preserves shape operators. If M is oriented by
the unit normal U, then since F* preserves dot products (and agrees with
F* on M), it follows that F*(£/) has unit length and is everywhere normal
to F(M) = M. Thus one of the unit normals on M, say U, has the property
that
F*(£/(p) = U(p) where p = F(p).
If S and S are the shape operators on M and M derived from U and U,
respectively, we shall show that
F*(£(v)) = £(F*(v)).
Again let a be a curve in M with initial velocity v. Thus F(a) is a curve
in M with initial velocity F*(v). If U is restricted to a, and U is restricted
toF(a), then F*(t/) = U (Fig. 6.25). Since F* preserves derivatives of
vector fields, we get
F*(S(v)) = F*(t/'(0)) = U'(0) = £(F*(v)).
But v and S (v) are tangent to M; hence F* may be replaced by F*. 
Our goal now is the converse of the preceding theorem, that is: If
M and M are isometric and have the same shape operators, then M and M
are congruent. This is the analogue of the basic result (Theorem 5.3 of
Chapter III) for curves. The condition M and M isometric corresponds to
the hypothesis that a and /? are unitspeed curves defined on the same
FIG. 6.25
Sec. 8]
CONGRUENCE OF SURFACES 299
interval, and, of course, "same shape operators" corresponds to
K = K, T = ±f .
8.2 Lemma Let F: M *• M be an isometry of oriented surfaces in E 3
that preserves shape operators (as in Theorem 8.1 ). Let E u E 2 be a tangent
frame field on M, with E u E 2 the transferred frame field on M . If E 3 _ and
E 3 are the unit normals orienting M and M, then E u E 2 , E 3 and E u E 2 , E z
are adapted frame fields on M and M. For the connection forms of these
frame fields, we have
F*{a>ii) = on (1 ^ *J ^ 3).
Proof. We already know from Lemma 5.3 that F*(wi 2 ) = W12. It remains
to prove that
F*(w i3 ) = &>a, fori = 1,2.
But Corollary 1.5 shows that this merely expresses the preservation of
shape operators in terms of connection forms. In fact, for j = 1, 2,
F*(ua){Ei) = 5x»(F*Ej) = S(F*^,)^ = F*(S{Ei))'F*{Ei)
= SEjEi = uaiEj).
Hence the forms F*(wi 3 ) and w t  3 are equal. I
8.3 Theorem Let F: M *• A^ be an isometry of oriented surfaces that
preserves shape operators; that is,
*V0S(v)) = S(F*(v))
for all tangent vectors v to M. Then M and M are congruent; in fact there
is an isometry F of E 3 such that F = F  M.
(If it should happen that
F* («(▼)) = 8(F*(y))
for all tangent vectors, then it suffices to reverse the orientation of either
M or M to get the hypothesis as stated.)
Proof. Fix a point p of M, and let E 3 and E z be the unit normals orient
ing M and M. By using Corollary 2.3 of Chapter III, it is easy to show
that there is a unique isometry F of E 3 which agrees with F at the selected
point p in the sense that
F(p) = F(p)
F* (v) = F* (v) for every tangent vector v to M at p
F*(£ 3 ( P )) = Et(F(p)).
We shall show that F is the required Euclidean isometry, in other words,
300 GEOMETRY OF SURFACES IN E 3 [Chap. VI
FIG. 6.26
that F(q) = F(q) for an arbitrary point q of M . Thus if a is a curve in M
from p to q, it suffices to prove that F(a) = F(a).
There is no loss of generality in assuming that a lies in the domain of an
adapted frame field E lf E 2 , E 3 on M. (If not, we could break a up into
segments for which this is true, and repeat essentially the following proof
for each segment in turn.) Our plan is to use the general criterion Theorem
5.7 of Chapter III to show that the curves F(a) and F(a) are identical.
A. The curve F(a). Restrict the frame field E u E 2 , E 3 to the curve a
(Fig. 6.26). Then by the connection equations,
El = V a ,Ei = 2 aaicDEj (1 ^ i ^ 3).
Now apply F* to this equation; since F* is linear and preserves derivatives,
we get
(F*E<)' = Z «*,(«') F* (Ej) (1 £ » £ 3). (Al)
iucts; hence F*E lf F*E 2 , F*E 3 is the frame field
Furthermore,
F(a)'.F*^ = d'Ei (A2)
Also F* preserves dot products; hence F*E lt F*E 2 , F*E 3 is the frame field
on the image curve F(a). Furthermore,
since
V{a)' = F*(«').
B. The curve F (a) = a. Use the isometry F to transfer the tangent
frame field E u E 2 to a tangent frame field E u E 2 on M. With the unit nor
mal vector field E z , we now have an adapted frame field E lf E 2 , E s
on M. We restrict it to the image curve F(a) = a, and use the connection
equations as above to get
El = Z Z>ij(a)Ej (1 ^ i ^ 3). (Bl)
Furthermore, we assert that
ol'Ei = cl*Ei (1 ^ i ^ 3). (B2)
For i = 1, 2 this follows immediately from the definition of Ei, E 2 , since
Sec. 8] CONGRUENCE OF SURFACES 301
F* is an isometry and
ol = F(a)' = F*(a).
For i = 3, both sides are zero, since a and 5 are curves in M and M , re
spectively.
C. Comparison of F(a) and F (a) = a. The construction above and the
assumption that F preserves shape operators has exactly reproduced the
hypotheses of the preceding lemma; hence F*(uij) — ua for 1 ^ i,j ^ 3.
Thus
««(»') = S><j(F*(d)) = (F*«<y)(«) = mi(d).
Using this fact we deduce from (Al) and (Bl) that
(F+EiY'F+Ej = Ei'E; (1 ^ ij ^ 3). (CI)
Comparing (A2) and (B2), we get
F(a)''F*Ei = a'Ei (1 ^ t ^ 3). (C2)
And by our initial construction we have
F*^i = Ei at the point p = a(0)
= F(a(0)) (1 ^ t ^ 3).
Refering to equations (J) in Theorem 5.7 of Chapter III, we observe
that the three equations (CI), (C2), (C3) are precisely what is needed
to conclude that F(a) = a; that is,
F(a) = F(a). I
This theorem gives a formal proof that the shape operators of a surface
M in E 3 do, in fact, completely describe its shape.
EXERCISES
1. A surface M C E 3 is rigid provided every surface isometric to M is
congruent to M . Deduce from Liebmann's theorem that spheres are
rigid.
2. If a, /8: / — * E 3 are unitspeed curves with K a = «/s > and t„ = rp,
show that their tangent surfaces are congruent. (Compare Ex. 5 of
VI.4.)
3. If M and N are congruent surfaces in E 3 , and F is a Euclidean isometry
such that F(M) = N, prove that F  M preserves Gaussian and mean
(C3)
302 GEOMETRY OF SURFACES IN E 3 [Chap. VI
curvature, principal curvatures, principal directions, umbilics, asymp
totic and principal curves, and geodesies. Which of these are preserved
by arbitrary isometries F: M —*■ N? (Hint: Orient locally, and ignore
ambiguity of signs for H, k ly and A; 2 .)
4. If F: 2 — ► 2 is an isometry of spheres, show that there is a Euclidean
isometry F such that F = F  2.
5. Let M be the saddle surface (z = xy). A rotation of 90° followed by
a reflection in the xy plane yields an orthogonal transformation C of
E whose matrix is
/0 1 0\
\0 1,
(With our conventions, the columns of the matrix are C(ui), C(ii2),
C(us), where u» is the ith. unit point.)
(a) Prove that C(M) = M.
(b) Let F = C  M: M > M. Orient M (as domain of F) by the unit
normal U such that £7(0) = u 3 . For which orientation of M (as
image of F ) does F preserve shape operators?
6. In the general description of a surface of revolution on page 129 (M
obtained by revolving C around A), let A be the line through p in
the direction of a unit vector ei, and let
a{u) = p + g(u) e x + h(u) e 2
be a parametrization of C, where e 2 is a unit vector orthogonal to d.
(a) Find a regular mapping x: D — » E whose image is the set M.
Then prove:
(b) M is congruent to a surface of revolution in the special position
given in Example 2.5 of Chapter IV.
(c) M is a surface in E .
(d) Two surfaces of revolution are congruent if and only if they can
be described in this way by the same pair of functions g, h.
7. If M is a surface in E 3 , a Euclidean isometry F such that ¥(M) = M
is called a Euclidean symmetry of M. Show that
(a) The set of all Euclidean symmetries of M forms a subgroup §>(M)
of the group S of all isometries of E 3 (Ex. 7 of III.l). §>(M) is
called the Euclidean symmetry group of M.
(b) The Euclidean symmetry groups of congruent surfaces are iso
morphic.
8. Show that the Euclidean symmetry group of any sphere is isomorphic
to the group of all 3 X 3 orthogonal matrices.
Sec. 9] SUMMARY 303
9. Find all eight Euclidean symmetries of the saddle surface M: z = xy.
Show that they are orthogonal transformations and give their matrices.
10. Find all Euclidean symmetries of the ellipsoid x 2 /a + y 2 /b 2 + z /c =1,
where a > b > c. (Hint: Use the fact that Gaussian curvature is
preserved.)
11. If M is Scherk's surface (Ex. 21 of V.4) and D is the open square
— 71/2 ^ u, v S t/2, show that:
(a) The image x(Z>) of x (Ex. 4 of V.4) lies in M.
(b) The portion of M over any open square (Ex. 21 of V.4) is con
gruent to x(D).
(c) The curvature formulas given in the abovementioned exercises
are consistent.
9 Summary
The geometrical study of a surface M in E 3 separates into three distinct
categories :
(1 ) The intrinsic geometry of M .
(2) The shape of M in E 3 .
(3 ) The Euclidean geometry of E 3 .
We saw in Chapters II and III that the geometry of E 3 is based on the dot
product and consists of all concepts preserved by isometriesof E 3 . Similarly,
we have now found that the intrinsic geometry of M is based on the dot
product— applied only to vectors tangent to M— and that it consists of
all concepts preserved by isometries of M .
The shape of M in E 3 is, in a sense, a link between these two geometries.
For example, Gaussian curvature K is an essential feature of the intrinsic
geometry of M, and the shape operator S dominates category (2) — thus
the equation
K = det S
shows that the geometries (1) and (3) can be harmonized only by means
of restrictions on (2). Stated bluntly: Only certain shapes are possible in
E 3 for a surface M with prescribed Gaussian curvature. A strong result of
this character is Liebmann's theorem, which asserts that a compact surface
in E 3 with K constant has only one possible shape — spherical.
In the last two chapters, the computation of explicit examples has been
mostly in terms of coordinate patches (Gauss), but the theory itself has
been expressed in terms of frame fields and forms (Cartan). Historically,
coordinates were used for the theory as well, but by now the Cartan ap
proach has largely won the day. We have seen in Section 6 that the two
approaches are not so far apart when the coordinate patch is orthogonal.
CHAPTER
VII
Riemannian Geometry
In studying the geometry of a surface in E 3 we found that some of its
most important geometric properties belong to the surface itself and not
the surrounding Euclidean space. Gaussian curvature is a prime example;
although denned in terms of shape operators, it belongs to this intrinsic
geometry, since it passes the test of isometric invariance. As this situation
gradually became clear to the mathematicians of the 19th century, Rie
mann drew the correct conclusion: There must exist a geometrical theory
of surfaces completely independent of E 3 , a geometry built from the start
solely of isometric invariants. In this chapter we shall give an outline
of the resulting theory, concentrating on its dominant features: Gaussian
curvature and geodesies. Our constant guides will be the two special cases
which led to its discovery: the intrinsic geometry of surfaces in E 3 , and
Euclidean geometry — particularly that of the plane E 2 .
1 Geometric Surfaces
The evidence from earlier work on the intrinsic geometry of surfaces in
E (and on Euclidean geometry as well) suggests that we will need the dot
product on tangent vectors to do geometry on a surface.
\But to free ourselves of confinement to E 3 , we must begin with an ab
stract surf ace M (Chapter IV, Section 8). Since M need not be in E 3 there
is no dot product — and hence no geometry. The dot product, however, is
but one instance of the general notion of inner product, and Riemann's
idea was to replace the dot product by a quite arbitrary inner product on each
tangent plane of M.
304
Sec. 1] GEOMETRIC SURFACES 305
1.1 Definition An inner product on a vector space V is a function which
assigns to each pair of vectors v, w in V a number vowso that these rules
hold:
(1) Bilinearity:
(fllVi + O2V2) o W = ttiVi o W + 02V2 o W
V o (61 Wi + 62W2) = 61V 0W1 + 62V o W2.
(2) Symmetry: vow = wov.
(3) Positive definiteness:
vov^ 0; and v o v = if and only if v = 0.
On the vector space E 2 the dot product
VW = V1W1 + V 2 W 2
is, of course, an inner product, but there are infinitely many others, such
as, for example, v o w = 2viWi + 3v 2 W2. (See Exercise 8 of Section 2.)
Shifting then from surfaces in E 3 to abstract surfaces and, from the dot
product to arbitrary inner products, we get the following definition.
1.2 Definition A geometric surface is an abstract surface M furnished
with an inner product, °, on each of its tangent planes. This inner product
is required to be differentiable in the sense that if V and W are (differentia
ble) vector fields on M, then V° W is a differentiable real valued function
on M.
We emphasize that each tangent plane T P (M) of M has its own inner
product, and they are unrelated save for the differentiability condition —
an obvious necessity for a theory founded on the calculus. In this definition
V°W has its usual pointwise meaning: It is the function on M whose
value at each point p is the number 7(p) ° W(p). An assignment of inner
products to tangent planes as in Definition 1 .2 is called a geometric structure
(or metric tensor or "ds ") on M.
In short :
Surface + geometric structure = geometric surface
and we emphasize that the same surface furnished with two different geo
metric structures gives rise to two different geometric surfaces.
1.3 Example Some geometric surfaces.
(1) The plane E 2 , furnished with the usual dot product on tangent
vectors, is the bestknown geometric surface. Its geometry is twodimen
sional Euclidean geometry.
(2) A simple way to get new geometric structures is to distort old ones.
306 RIEMANNIAN GEOMETRY [Chap. VII
For example, if g > is any differentiable function on the plane, and • is
the usual dot product, define
Vo W
9 2 (p)
for tangent vectors v and w to E 2 at p. This is a new geometric structure
on the plane, said to be conformal to that of the dot product (Exercise 1).
We shall see that (unless g is special) the resulting geometric surface has
properties quite different from the Euclidean plane (1 ) .
(3) If ¥ is a surface in E , then the dot product from E 3 applied to
tangent vectors on M furnishes an inner product making M a geometric
surface. This, of course, is just what was done in Chapters V and VI. Unless
some other inner product is explicitly mentioned, it is always assumed that
a surface in E is made geometric in this way.
Here a word about terminology is in order. Euclid's name carries geo
metric implications. Hence in Chapter I, where geometry did not appear,
we should have called E 2 the Cartesian plane, reserving the term Euclidean
plane for the geometric surface (1 ) above.
From the simple beginning in Definition 1.2, it is rather surprising what a
rich geometric theory can be built. But as mentioned earlier, examples
(1) and (3) certainly indicate that the theory is there to be explored,
and their common features even suggest the kind of results we may expect
to find.
The definitions in Chapter VI that are clearly intrinsic in character
will be used here without further discussion. In particular, an isometry F:
M — > M of arbitrary geometric surfaces is exactly as denned in Definition
4.2 of Chapter VI and the geometry of M consists by definition of its iso
metric invariants. A frame field on an arbitrary geometric surface M con
sists, as usual, of two orthogonal unit vector fields E\, Ei defined on some
open set of M. The orthonormality equations
EioEj = Sn (1 ^ i,j ^ 2)
are expressed, of course, in terms of the inner product of M . As before,
we derive the dual 1forms 0i, 02, characterized by di(E,) — b ijy and then
the connection form un = — «2i, characterized by the first structural
equations,
ddi = C0]2 A 02, ddz = G>21 A 01
We emphasize once more that these forms 0i, 02, un are not invariantly
attached to the geometric surface M; a different choice of frame field
Ei, Ei will produce different forms 0i, 2 , wi 2 . Before going any further, we
had better see how two such sets of forms are related.
Sec. 1]
GEOMETRIC SURFACES
307
E t
E,
*~E t
\
(A)
FIG. 7.1
On a small enough neighborhood of a point p,
careful use of the inverse function cos  (or sin ) E t
will yield a differentiate function <p such that \^
Ei = cos (p Ei + sin <p E 2 .
We call <p an angle function from Ei, E 2 to Ei, E 2 .
As shown in Fig. 7.1, there are now two possibili
ties for E 2 . Either
Ei = — sin <p Ei + cos <p E 2
in which case we say that Ei, E 2 and E lf E 2 have the
same orientation, or
E 2 = sin <p Ei — cos <p E 2 ,
which is opposite orientation.
1.4 Lemma Let E\, E 2 and E u E 2 be frame fields on the same region in
M. If these frame fields have
(1) The same orientation, then
W12 = cow + d(p, and 0i a s 2 = 0i a 6 2 .
(2) Opposite orientation, then
«i2 = — (o>i2 + rf^), and 0i a 2 = — 1 a 2 .
Proof. We discuss only the first case, since the second is obtained from
it by merely changing signs. By the basis formulas (Lemma 2.1 of Chapter
VI), the equations
Ei = cos <p Ei + sin <p E 2 , E 2 = —sin <p E x + cos ^ E 2
yield
0i = cos <p 0i — sin £> 2 , 02 = sin <e> 0~i + cos ^02. ( )
Applying the exterior derivative to the first of these, we get
rf0i = — sin <p d<p a 0i + cos <p ddi — cos <p d<p A d 2 — sin <p dd 2 .
Now we substitute the first structural equations for ddi, dd 2 to obtain
ddi = (wi2 — d(p) a (sin <p 0i f cos ^> 2 )
= (<ii2 — d<p) a # 2 .
In the same way we get
dd 2 = — (a>i2 — dip) a 0j.
Because the form a>i2 = — co 2 i uniquely satisfies the first structural equa
tions, we conclude from the last two equations that coi2 = 0012 — dtp, as
308
RIEMANNIAN GEOMETRY
[Chap. VII
FIG. 7.2
required. Direct computation of 0i a 2 using (*) above shows that this
2form equals 0~i a 2 . 
The concept of geometric surface can be used to fill a gap in our earlier
work. Occasionally we met regular mappings x: D —> E 3 which were not
parametrizations of any surface in E 3 . For example, the image x(Z>) of D
might fold back through itself as indicated in Fig. 7.2, so that the definition
of surface in E fails along the crossing line L.
This technical difficulty can be eliminated by assigning D (which is a
surface) not the usual dot product, but instead the induced inner product
vow= x*(v)«x*(w).
Thus D becomes a geometric surface, and if x(D) were a surface in E , x
would evidently be an isometry. In short, D has exactly the intrinsic geome
try we might intuitively expect x(D) to have.
In this chapter, as earlier, the restriction to low dimensions is not essen
tial. A surface is the twodimensional case of the general notion of mani
fold (Chapter IV, Section 8). A manifold M of arbitrary dimension fur
nished with a (differentiable) inner product on each of its tangent spaces
is called a Riemannian manifold, and the resulting geometry is Riemannian
geometry. (Euclidean geometry, as discussed in Chapter III, is the special
case of Riemannian geometry obtained on the Euclidean space E™, with
its usual dot product.) A geometric surface is thus the same thing as a two
dimensional Riemannian manifold, and the subject of this chapter is two
dimensional Riemannian geometry, f
EXERCISES
1 . For a conformal geometric structure on the plane (Example 1 .3 ) , show
that
tWe would prefer to call a geometric surface a Riemann(ian) surface but this
term has a firmly established and distinctly different meaning.
Sec. 1] GEOMETRIC SURFACES 309
(a) The formula
vow =  v   w  cost? (where  v  = y/v°v)
gives the same value for the angle ^ # ^ t between v and w as
in the Euclidean plane E 2 .
(b) The speed of a curve a = (an, a 2 ) is y/a'i 2 + a 2 2 /g(a).
(c) gUi, gU 2 is a frame field with dual forms du/g, dv/g.
(d) The area forms are ± dudvl g 2 .
Note that g = 1 gives the usual Euclidean structure.
2. The Poineari halfplane is the upper halfplane v > furnished with
the inner product (o) obtained by dividing the dot product at each
point p by the square of the distance v(p) = p 2 to the u axis:
vow = v • w/y (p ) .
For the curve a{t) = (r cos t, r sin t), < t < w, find the speed and
arclength function (measure from the top of the semicircle, t = x/2. )
3. (a) On a geometric surface M, let V and W be vector fields that are
linearly independent. Find a frame field for which Ei = V/\\ V .
(b) Deduce an explicit formula for a frame field on the image of an
arbitrary patch xinM.
4. If dM is an area form on M and v is an arbitrary 2form, show that
there is a function / such that v = f dM. Deduce that a (connected)
orientable geometric surface has exactly two area forms, ± dM.
5. Let M be a geometric surface oriented by area form dM. Prove
(a) On each tangent space to M there exists a unique "rotation by
+90°," that is, a linear operator J: T P (M) * T P (M) such that
ll^(v) = v, J(v)° v = 0,
and
dM(v,J(y)) > (if v ^ 0).
(Hint: If E u E 2 is a positively oriented frame field,
J(E,) = E 2 , J(E 2 ) = El)
We call these operators — collectively, for all points of M — the
rotation operator^ of M .
(b) J is differentiate (J(V)° J(W) is differentiate for any vector
fields V, W) and skewsymmetric
J{V)°W + V°J(W) = 0,
and J 2 = —I (J applied twice is minus the identity operator).
t Compare the special case in Exercise 14 of Section 4, Chapter II.
310 RIEMANNIAN GEOMETRY [Chap. VII
(c) If M is oriented instead by —dM, then its rotation operator is —J.
(d) If M is a surface in E and its orientation is given by the unit
normal U, then J(V) = U X V.
The operator J serves as a kind of replacement for the unit normal
on surfaces that are not in E 3 . In particular, (d) shows how the scheme
given in Chapter VI (page 291) for measuring oriented angles applies
now to an arbitrary (oriented) geometric surface.
6. If F: M — > N is a regular mapping of oriented geometric surfaces, show
that the following are equivalent:
(a) F is orientationpreserving and conformal (Definition 4.7 of
Chapter VI).
(b) F preserves the rotation operators of M and N; that is,
F*(J(v)) =J(F,(v))
for all tangent vectors v to M .
(c) F preserves oriented angles; that is, if # is an angle from v to w,
then # is also an angle from F* (v) to F* (w).
7. (a) Prove that a regular mapping F = (/, g): E 2 — * E 2 is orientation
preserving and conformalf if and only if f u = g v , f v = —g u .
If E 2 is considered as the complex plane, with z = u + iv = (u,v),
these two equations (the CauchyRiemann equations) are necessary and
sufficient for F to be a complex analytic function z —> F(z).
(b) Given such a complex function F, show that its scale factor X(z) is
the magnitude of the (complex) derivative dF/dz.
8. If the origin is deleted from E 2 , show that the mapping F in (2) of Ex
ample 7.3 of Chapter I is orientationpreserving and conformal. What
is the complex function in this case?
9. Let D and E be regions in the plane, furnished with conformal geometric
structures given by functions g\ and gi , respectively. Let D' and E'
be the same regions, with the usual Euclidean structure. If F: D' — » E'
is a conformal mapping with scale factor X, prove that F: D — > E is
conformal with scale factor \gi/gi(F).
2 Gaussian Curvature
For arbitrary geometric surfaces, we need a new definition of Gaussian
curvature. The definition K = det S for surfaces in E is meaningless now,
t The term conformal is often understood to include preservation of orientation.
Sec. 2] GAUSSIAN CURVATURE 311
since it is based on shape operators. But this original definition made K
an isometric invariant, so it is reasonable to look to the proof of the theo
rems egregium (specifically to Corollary 2.3 of Chapter VI) to find a satis
factory generalization.
2.1 Theorem On a geometric surface M there is a unique real valued
function K such that for any frame field on M the second structural equation
dun = —Kdi A 2
holds. K is called the Gaussian curvature of M.
Proof. For each frame field E u E 2 , there is (by the basis formulas of
Lemma 2.1 of Chapter VI), a unique function K such that
d<t3i2 = —Kdi a 2 .
But another frame field E u E 2 might a priori have a different function
K such that
du l2 = Kdi A 02
What we have to show is consistency: Where the domains of these frame
fields overlap, there K = K. Since such domains cover all of M (Exercise
3 of Section 1), we will then have a single function K on M with the re
quired property. This consistency will follow immediately from Lemma 1.4.
First consider the case in which the frame fields have the same orienta
tion, so u n = a>i2 + d<p. Hence dun = dun, because d = 0. But then
Since
we conclude that
Kdi A 62 = Kd\ A 02.
01 A 2 = di A 2 y£ 0,
K = K.
When the orientations are opposite, we get dun = —dun but still find that
K = K, since
01 A 2 = 0 X A 2 . 
As noted above, Corollary 2.3 of Chapter VI shows that this general
definition of Gaussian curvature agrees with the definition K = det S when
M is a surface in E 3 . The proof of isometric invariance obtained there is
entirely intrinsic in character, and it thus holds for arbitrary geometric
surfaces.
Gaussian curvature is the central property of a geometric surface M ;
312 RIEMANNIAN GEOMETRY [Chap. VII
it influences — often decisively — many of the most important properties
of M. In Section 6 we shall examine the influence of curvature on geodesies
and in Section 8 its influence on the topology of M .
To summarize: Geometrical investigations in terms of a frame field
Ei, E 2 are dominated by its structural equations:
iddi = a>i2 a Bi,
\ddi = C021 a S\,
dun = — Kdi a 2 .
The first structural equations actually define the connection form
0>12 = — C021
of that frame field, while the second structural equation defines the Gaussian
curvature K of the geometric surface (independent of choice of frame field ) .
It is already clear from Chapter VI, Section 6, how oi 12 and K may be ex
plicitly computed from these implicit definitions.
2.2 Example Gaussian curvature.
(1) The Euclidean plane E . If we use the natural frame field Ui, U 2 ,
then the dual 1forms are merely d x = du, 2 = dv. Since dd\ = dd 2 = 0,
the identically zero form o>i 2 = satisfies the first structural equations and
hence is the connection form of Ui, Ui. But then dun = 0, so K = 0.
The Euclidean plane is flat. This can be no surprise, since E 2 is isometric
to a plane in E , for which we know that K = 0, since its shape operators
all vanish.
(2) The plane with conformal inner product
v • w
V o w =
<7(p) 2
(See (2) of Example 1.3.)
The natural Euclidean frame field U\, Ui is no longer a frame field rela
tive to this new inner product. U\ and Ui are still orthogonal, but
UtoTJi = U 2 oU 2 = ,.
9 2
Thus gU\, gUi is a frame field. It follows easily that its dual 1forms are
a  du a  dv
V\ — , C/ 2 •
g 9
To find the connection form W12, we first differentiate 9 X and Q%.
Sec. 2] GAUSSIAN CURVATURE 313
ddi = d (  J A du
dd 2 = d (  J a dv.
Now d (l/g) = —dg/g 2 and dg = g u du + £» efo. Because dw dw = dvdv = 0,
we find
ddi = ( 9 ~ du) a 2
/ \ ()
dd 2 = l^dvj a Bx.
Comparison with the first structural equations then yields
wis =  (ft du — g u dv)
9
because by (*), this form coi 2 satisfies the first structural equations; by
uniqueness it must be the connection form.
To get the curvature we differentiate once more.
duu = d I  ) a (g v du — g u dv) +  (g™ dv du — g uu du dv) .
W 9
From the above we know that
d ( ) = — r (g u du + g, dv)
\9/ 9 2
and that du = gdi, dv = gd 2 . A simple computation using these facts then
gives
dwi2 = (g u 2 + gv — g(guu + g™)) fa A ^a
Thus by the second structural equation we conclude that
K = g(g uu + gw)  (gu + gv).
The induced inner product discussed on page 308 may be applied in
other situations. For example, suppose that F: M — > N is a diffeomorphism
of surfaces (Chapter IV, Section 5) and that N is a geometric surface.
Then the induced inner product
vow = F* (v) oF* (w)
on tangent vectors to M , makes M a geometric surface — and F an isome
try. M might be called a "new model" of N; however different it may look,
it is geometrically identical with AT.
314
RIEMANNIAN GEOMETRY
[Chap. VII
FIG. 7.3
2.3 Example
(1) The stereographic sphere. We have proved in Example 5.5 of Chapter
IV that stereographic projection P is a diffeomorphism of the punctured
sphere 2 onto the Euclidean plane E 2 . Now consider S merely as a surface,
while E is geometric, with its usual dot product. Thus the induced inner
product makes S a geometric surface which is isometric to E 2 and hence
flat. If 2 appears round, it is only because we look at it with Euclidean
eyes; that is, we erroneously assume it must have the dot product of E 3
as in Chapter V.
(2) The stereographic plane. Now let us reverse the process in (1 ) . Con
sider 2 with its usual geometric structure as a surface in E 3 , and let E 2 be
merely a surface.
The inverse P _1 : E 2 — > S of stereographic projection is also a diffeo
morphism. The inner product (°) induced by P _1 on E 2 makes E 2 a geo
metric surface (the stereographic plane) isometric to S , and thus having a
curvature K = +1.
We examine this new stereographic plane more closely. Throughout
a dot (• ) will as usual denote the dot product, whether of E 2 or E 3 .
If v and w are tangent vectors to E 2 at q = P(p), let v and w be the
unique tangent vectors to 2 at p such that P* (v ) = v, P* (w) = w (Fig.
7.3). Now by Exercise 14 of Section 4, Chapter VI, we know that
vw = P*(v) • P*(w)
+ w
But (P *)* carries v and w back to v and w, so for the induced inner
product of E 2 we find
vo W = (P X )*(v) • (P 1 )*(w) = v«w
 + ¥r
It follows immediately that this inner product is of the conformal type
discussed in Example 1.3 with
1 +
2 I I
u + v
Sec. 2] GAUSSIAN CURVATURE 315
To visualize this unusual "plane," we may imagine that rulers get longer
as they move farther away from the origin. Since P is now an isometry, the
intrinsic distance from p to q (in Fig. 7.4) is exactly the same as the dis
tance from p* to q* . Also circles u 2 + v 2 = r 2 , for which r is very large,
actually have very small stereographic arc length since they correspond
(under the isometry P) to small circles about the north pole in S .
2.4 Example The hyperbolic plane. Let us experiment with a change
of sign in the stereographic inner product above, setting
2 , 2
! U + V
Since g > is necessary, this hyperbolic inner product vow = (l/g )v»w
is used only on the disc u 2 + v 2 < 4 of radius 2 in the plane. The resulting
geometric surface is called the hyperbolic plane H.
In this case
_ ~ u _ ~~ v a — — — *
9u n~ , gv — ~n~ ■> ancl g U u — ff™ — — tt >
so the general computations in Example 2.2 show that
wi2 = — (u dv — v du)
and that the hyperbolic plane has constant Gaussian curvature K = —1.
As a point (u, v) approaches the rim of H, that is, the circle u 2 + v =4
(not a pait of HI), g(u, v) approaches zero. Thus in the language used
above, rulers must shrink as they approach the rim, so that H is a good
deal bigger than one's Euclidean intuition may suggest. For example,
for # constant let us compute the arc length function s(t) of the Euclidean
line segment
a(t) = (t cos t», t sin 0), S t < 2,
which runs from the origin almost to the rim. Now a = (cos t?, sin#),
so a °a' = \/g(a) 2 . But
316 RIEMANNIAN GEOMETRY [Chap. VII
g(a(t)) = 1,
so a has hyperbolic speed
II «'(«) II =
1
g(a) (1  t 2 /4)
Thus
s(() = rr^k = 2tanh "1 = ^
2 + <
f 2/4  " ""— 2 ~ iWfe 2  « *
Thus as t approaches 2, arc length s(t) from the origin a(0) to «(£)
approaches infinity. This "short" segment a actually has infinite hyperbolic
length. Further properties of the hyperbolic plane will be developed as we
go along. We shall see that it — and not the bugle surface (Example 6.6 of
Chapter V) — is the true analogue of the sphere for constant negative
curvature.
2.5 Example A flat torus. Let T be a torus of revolution considered
merely as a surface, and let x: E 2 — » T be its usual parametrization (Exam
ple 2.6 of Chapter IV). Now we give T a geometric structure by defining
It is easy to check that this defines — without ambiguity — an inner product
on each tangent plane of T.
Because x*(?7i) = x„ and x*(Uz) = x„, it follows immediately that
x is a local isometry of the Euclidean plane E 2 onto the geometric surface
T. Since local isometries also preserve Gaussian curvature, T is flat . Thus
its geometric structure is different from the usual torus in E , which has
variable curvature.
Because this torus T is compact and flat, Theorem 3.5 of Chapter VI
shows that it can never be found in E . Explicitly, there exists no surface
M in E 3 that is isometric to T, for then M would also be compact and flat —
but this is forbidden by the theorem. Here we have a proof that the class
of geometric surfaces is richer than that of surfaces in E . In the course of
this chapter we hope to convince the reader that geometric surfaces are
the natural objects to study and that surfaces in E — however intuitive
they may seem at first glance — are no more than an interesting special case.
One should not conclude from the example above that every surface
can be given a flat geometric structure. Topological subtleties are involved,
as we shall see in Section 8.
2.6 Remark So far we have reserved the dot notation (•) for the dot
product of Euclidean space, and used a small circle (°) to emphasize the
Sec. 2] GAUSSIAN CURVATURE 317
generality of the inner product of an arbitrary geometric surface. From
now on we shall use a dot for all inner products, reverting to the former
convention only when, as in Example 2.4, the two appear in the same
context.
EXERCISES
1. Derive the dual forms and connection form «i 2 = du/v for the frame
field vUi, vU 2 on the Poincar6 halfplane (Ex. 2 of VII.l) and show
that this surface has constant negative curvature K = — 1.
2. For the conformal geometric structure on the entire plane with
g = cosh (uv),
compute the dual forms and connection form of the frame field gU u
gUi, and derive the Gaussian curvature K.
3. Find the area A of the disc u 2 f v S r 2 in the hyperbolic plane. (Hint
Find E, F, G for a 2segment
x(w,v) = (u cos v, u sin v).)
What is the area of the entire hyperbolic plane?
4. The hyperbolic plane of pseudoradius r is obtained by altering the
function g in Example 2.4 to g = 1 + (u + v)/Ar 2 . Find its Gaussian
curvature.
5. Find the area of the flat torus in Example 2.5. Modify the definition
so as to produce a flat torus with arbitrary area A > 0.
6. Show that there is a geometric structure on the projective plane such
that the natural mapping P: S — * S is a local isometry. Prove that this
geometric surface 2 cannot be found in E 3 . (The same results hold when
S is a sphere of radius r and S becomes the projective plane of radius r.)
7. Show that the plane, furnished with the conformal geometric structure
such that g = sech u, is isometric to a helicoid.
8. The identity map \(u,v) = (u,v) of E 2 is a patch with x„ = Ui,x v = Ui.
Thus if (° ) is a geometric structure on the plane, this patch has
E = Ui o U u F = U 1 o U 2 , G = Ui o Vi.
(a) Given any differentiate functions E, F, and G on the plane, such
that E > 0,G> 0, EG — F 2 > 0, show that there is a geometric
structure on the plane corresponding, as above, to these functions.
(b) Show that the method used in Example 2.2 is a special case of
318 RIEMANNIAN GEOMETRY [Chap. VII
that of Chapter VI, Section 6, and derive the formula for K in
Example 2.2 from Lemma 6.3 of Chapter VI.
{Hint: For (a) define
vow = EviWi + F(viW 2 + v 2 Wi) + Gv 2 w 2 .)
3 Covariant Derivative
The covariant derivative V of E 3 (Chapter II, Section 5) is an essential
part of Euclidean geometry. We used it, for example, to define the shape
operator of a surface in E , and in modified form (Chapter II, Section 2)
to define the acceleration of a curve in E 3 . In this section we will show that
each geometric surface has its own notion of covariant derivative.
As in Euclidean space, a covariant derivative V on a geometric surface M
assigns to each pair of vector fields V, W on M a new vector field V r W,
and we must certainly require that it have the usual linear and Leibnizian
properties (Corollary 5.4 of Chapter II). Intuitively, the value of V Y W
at a point p will be the rate of change of W in the V (p) direction. Thus if
the connection form uu of a frame field E\, E 2 is to have its usual geometri
cal meaning (measuring rates at which E\ turns toward E 2 ), we must also
require that
co 12 (7) = V v EiE 2 . (*)
These conditions completely determine V V W for any vector fields V and W:
3.1 Lemma Assume that V is a covariant derivative on M with the usual
linear and Leibnizian properties, and such that (*) holds for a frame
field Ei,E 2 . Then V obeys the connection equations
V v Ei = o) n (V)E 2
V V E 2 = (aa(V)Ei.
Furthermore if W = fiEi + fiE?. is an arbitrary vector field, then
V r W = {V[fi] +/ 2 co 21 (F)}£ 1 + {V\M +/ 1 « a (F)}^,.
We call this last expression the covariant derivative formula. Note that
F[/i] and V\fi] only tell how W is changing relative to Ei,E 2 — the effect of
the terms involving connection forms is to compensate for the way E x , E 2
itself is rotating, so V Y W is an "absolute" rate of change.
Proof. Since Ei'E 2 = 0, by a Leibnizian property of V, we get
= V[.£a*.E(2] = VyE\»E 2 f E\ m ^yE 2 .
Sec. 3] COVARIANT DERIVATIVE 319
Hence, by (*),
VyE 2 »E 1 = wuOO = ton (7).
But Ei*Ei = 1, so the same Leibnizian property gives 2V Y Ei*Ei = for
i =■ 1, 2. Using this information, we derive the connection equations by
orthonormal expansion of V v Ei and V V E 2 .
Finally we apply the assumed properties of V to get
= 7[/lR + /l V V ^! + 7[/ 2 ]#, + /2 V v # 2 .
Substitution of the connection equations then gives the covariant deriva
tive formula. I
This lemma shows how to define the covariant derivative of M. Note that
the order of events is the reverse of that in Chapter II. There we used the
Euclidean covariant derivative to define connection forms; now we shall
use the connection form o>i 2 to define the covariant derivative for M.
3.2 Theorem For a geometric surface M, there is one and only one co
variant derivative V with the usual linear and Leibnizian properties
(Corollary 5.4 of Chapter II) and satisfying equation (*) for every frame
field on M.
Proof. The previous lemma shows that there is at most one such co
variant derivative, for V r W is given by a formula which does not involve
V. So what we must prove is that such a covariant derivative V actually
exists. The proof split into two parts, and we shall supress some details.
A. Local definition. For a fixed frame field E lf E 2 on a region 0, use the
formula in Lemma 3.1 as the definition of V r W. Routine computations
verify that V is linear and Leibnizian, and when W is E lt we get
V r Ei = un(V)E 2 ;
hence (*) holds.
B. Consistency. For two different frame fields, do the local definitions
agree? If V V W derives from E u E% on 0, we must show that V r W = V V W
holds on the overlap of and 0. For then we have a single covariant de
rivative on all of M. Because of the linear and Leibnizian properties, it
suffices to show that
V Y Ei = V F Bi, VvE* = vVl?2. (1)
We use Lemma 1.4, assuming for simplicity that E u E 2 and E u E 2 have
the same orientation. Applying VV to the equation
320 RIEMANNIAN GEOMETRY [Chap. VII
E t = cos # Ei + sin # # 2 ,
the covariant derivative formula gives
V v Ei = { F[cos #] + sint? co2i(F)}^i + { 7[sin#] + cost? o> n {V)}E 2 . (2)
By Lemma 1.4, a>i 2 = coi 2 + d#. Substituting w 12 = a>i 2 — d& into (2)
produces some favorable cancellations, leaving
V Y E\ = a>a(V){ — sin & Ei + cos # # 2 }
_   (3)
= &i 2 (V)E 2 = VyE X .
In the same way, V r E 2 = W#2 may be derived from
Ei = —sin # Ei + cos # Z? 2 . 
3.3 Example The covariant derivative of E 2 . The natural frame field
Ui, U 2 has coi2 = 0. Thus for a vector field
W = hUx + ftU lf
the covariant derivative formula (Lemma 3.1) reduces to
V r W = V\Wi + V[f 2 ]U 2 .
This is just Lemma 5.2 of Chapter II (applied to E 2 instead of E ),
so our abstract definition of covariant derivative produces correct results
on the Euclidean plane.
The covariant derivative V of a geometric surface M may be modified
so as to apply to a vector field Y on a curve a in M . (As usual, for each t,
Y (t) is a tangent vector to M at a (t), as in Fig. 7.6. )
If Ei, Ei is a vector field on a region of M containing a, we may write
Y(t) = y!(0^i(«(0) + y*(t)E,(a(t)),
or, briefly,
Y = yiEi + y 2 E 2 .
Roughly speaking, we want the covariant derivative Y of Y to be V a ,Y.
Thus the covariant derivative formula (Lemma 3.1) shows that we must
define
Y' = {yi + yxatiict )\Ei + {y 2 + yiun (a ))E 2 .
It is a routine matter to check that this notion of covariant derivative is
independent of the choice of frame field and has the same linear and
Leibnizian properties as in the Euclidean case. Also, as in Chapter II, Sec
tion 2, since the velocity a of a curve in M is a vector field on M, we can
take its covariant derivative to obtain the acceleration a." of a.
Sec. 3] COVARIANT DERIVATIVE 321
FIG. 7.5
It may be well to look back now at the case of a surface M in E . If V
and W are tangent vector fields on M, there are two ways to take covariant
derivatives: one from the intrinsic geometry of M as a geometric surface,
the other the Euclidean covariant derivative of E 3 . These two derivatives
are generally different, but there is a simple relationship between them.
3.4 Lemma Let V and W be tangent vector fields on a surface M in E 3
(Fig. 7.5). If V is the covariant derivative of M as a geometric surface,
and V is the Euclidean covariant derivative, then
V r W is the component of y r W tangent to M.
Proof. First suppose that W is one of the vector fields E lf E 2 of an adapted
frame field E u E 2 E 3 . By the Euclidean connection equations (Theorem
7.2 of Chapter II) we have
vY#i = E <*a<y)Ei = <»n(V)E 2 + a> 13 (F)#3.
3=1
But the connection equations (Lemma 3.1) for M give
VyEi = 0> 12 (V)E 2 .
Thus VvEi is V v Ei plus a vector field normal to M. In other words, V v Ei
is (at each point) the component of VrEi tangent to M. The same result,
of course, holds for E 2 .
In the general case, since W is tangent to M, we may write
W = fiE, + ftEt.
Then the required result follows immediately from the special case above,
since both covariant derivatives are linear and Leibnizian. 
Thus we have been using the intrinsic covariant derivative of McE
all along, although we did not give it formal recognition. It occurs when
322 RIEMANNIAN GEOMETRY [Chap. VII
7(0) = V~ ^.p  a(0)
FIG. 7.6
ever we take the tangential component of the Euclidean covariant de
rivative.
Only the most basic properties of covariant derivatives are shared by
all geometric surfaces. In particular, the related notion of parallelism
(due to Levi Civita) does not always behave as in the Euclidean case.
Much of the individual character of Euclidean geometry rests on the fact
that a tangent vector \ p to, say, E 2 may be moved to a parallel tangent
vector v<j at any other point q. As we shall see, this phenomenon of "distant
parallelism" does not obtain on an arbitrary geometric surface. However
it is always possible to define parallelism of a vector field Y on a curve. In
Euclidean space, this means that Y has constant coefficients relative to the
natural frame field, but the infinitesimal characterization Y = makes
sense in general.
3.5 Definition A vector field Y on a curve a in geometric surface M is
parallel provided its covariant derivative vanishes : Y' = 0.
Just as in the Euclidean case, a parallel vector field has constant length,
for  Y  2 = YY, and (Y*Y)' = 2YY f = 0.
3.6 Lemma Let a be a curve in a geometric surface M , and let v be a
tangent vector at, say, p = a(0). Then there is a unique parallel vector
field Fona such that 7(0) = v (Fig. 7.6).
Proof. We may suppose that a lies entirely in the domain of a frame
field Ei, Ei on M. (Otherwise we could break a up into segments for which
this is the case.) The vector field 7 must satisfy the conditions
V' = 0, 7(0) = v. (1)
Because 7 has constant length  7  = c, we may write
V = c cos <p Ei + c sin <p E 2 (2)
where <p is the angle from E x to 7. Thus the covariant derivative formula
gives
Sec. 3] COVARIANT DERIVATIVE 323
V' = c { — sin <p <p + sin <p co2i(a )} E x
+ c {cos <p <p + cos <e> coi 2 (a )} #2
It follows immediately that (1) is equivalent to
<p = —oon(a ),
with^>(0) the angle from E'i(p) to V(0) = v. There is only one such func
tion, namely
<p(t) = (p(0) — I unia ) dt
Jo
This function <p, substituted in (2), defines the required vector field V. 
In the situation stated in Lemma 3.6 we say, for each t, that the vector
V(t) at a(t) is obtained from v at p = a(0) by parallel translation along a.
In E 2 , parallel translation of a tangent vector y p along a curve segment
from p to q merely produces the distantparallelism result v q , which is
thus entirely independent of the choice of the curve. But for an arbitrary
geometric surface M, different curves from p to q will usually produce
different vectors at q. Equivalently: // a vector vatpis paralleltranslated
around a closed curve a (starting and ending at p) the result v is not neces
sarily the same as v. This phenomenon is called holonomy. If we fix a par
ticular frame field on the curve a, then the proof of Lemma 3.6 shows that
parallel translation from a (a) = p to a(b) = p along a rotates all vectors
through the same angle <p(b) — <p(a) — since <p is the same for all parallel
vector fields. We call this the holonomy angle yp a of a. (Multiples of 2r
may be ignored in \f/ a , since they do not affect the determination of v*.)
3.7 Example Holonomy on the sphere 2 of radius r. Suppose the closed
curve a parametrizes a circle on 2. There is no loss of generality in assum
ing that a is a circle of latitude, say the wparameter curve
a(u) = x(u,v ), ^ u ^ 2x,
where x is the geographical patch in 2 (Fig. 7.7). According to Example
6.2 of Chapter VI, the associated frame field E u J£ 2 of x has o>n = sin v du.
Now the proof of Lemma 3.6 shows that every parallel vector field on a
has angle p (measured from #1) satisfying <p — — com (a ). It follows that
<p has constant value —sin vq on a. Thus the holonomy angle \f/ a of a is
<p(2ir) — <p(0) = — 2k sin v
Note that only on the equator, v = 0, does a vector v return to itself
after parallel translation around a. When Vo is near ir/2, then a is a small
circle around the north pole of 2. Since <p is close to —1, parallel vector
field V is rotating rapidly with respect to Ei, E 2 . But the holonomy angle is
324
RIEMANNIAN GEOMETRY
[Chap. VII
T,(2)
FIG. 7.7
close to — 2t, so the actual difference between v = V(0) and v* = V(2ir)
is, as one would expect, quite small.
Gaussian curvature has a strong influence on holonomy, as shown by
Exercise 5.
For a patch x in an arbitrary geometric surface M we shall inevitably
use the notation x uu for the covariant derivative of x u along the wparame
ter curves of x — with corresponding meanings for x uv , x„, and x vv . Thus
when M is a surface in E , we must have a new notation, say x.
, for
the analogous objects defined in Chapter V, Section 4. There we were
using the covariant derivative of E 3 , while now we use the covariant de
rivative of M itself. It is still true in the intrinsic case that x uv = x vu ,
but the proof is by no means obvious (Exercise 9 ) .
EXERCISES
1. In the Poincare' half plane, let a be the curve given in Exercise 2 of
Section 1. Express its velocity and acceleration in terms of the frame
field
Ei = vUi, E 2 = vU 2 .
(Hint: They are collinear.)
2. Let j8 be the curve
8(0 = (ct,st),t > 0,
in the Poincare" half plane, where c and s are constants with c 2 + s 2 = 1.
(Thus /3 is a Euclidean straight line through the origin.) Express the
velocity and acceleration of in terms of the frame field used in Exer
cise 1.
3. If V and W are tangent vector fields on a surface in E 3 , refine the proof
of Lemma 3.4 to show that
Sec. 3] COVARIANT DERIVATIVE 325
V V W = VyW + S(V)WU,
where S is the shape operator derived from U = ztE 3 . Hence if a is a
curve in M,
a" = a." + S(a)a U.
4. Show that on the sphere 2 the curve a given in Example 3.7 has (in
trinsic) acceleration a" = r cos Vo sin vo E 2  Compute its Euclidean
acceleration, and show that a" is the component tangent to 2.
5. Let a be a closed curve in a geometric surface M.
(a) If a is homotopic to a constant via x (Ex. 12 of IV.6), show that
the holonomy angle of a is J/ x K dM. (Assume that x(R) lies
in the domain of a frame field.) When x is patchlike, this inte
gral is the total curvature of the region x(R.)
(b) Compute the holonomy angle in Example 3.7 by this method.
6. Let V be a parallel vector field on a curve a in M, and let W be a vector
field on a with constant length. Show that W is parallel if and only if
the angle between V and W is constant.
7. Show that isometries preserve covariant derivatives in this sense: If
Y is a vector field on a curve a in M, and F: M — * M is an isometry,
then
F* (Y f ) = ?', where Y is the vector field F* (F) on a = F (a) in Af.
(Simplify matters by assuming that Y can be written Y = fE u where
Ei, Ei is a frame field on M).
This is the analogue of the Euclidean result (Corollary 4.1 of Chapter
III). The general case is given in Exercise 8.
8. Prove that an isometry F.M+M preserves the covariant derivatives
V and V. Explicitly, for each vector field X on M , let X be the trans
f ered vector field o n M: X(F(p)) = F*(X (p )) for each point of M .Then
show that VyW = V#. (Hint: If /< = W *Ei and /» = WE h use Exer
cise 8 of Section 5, Chapter IV, to show that V[fi] at p equals V[fi] at
F(p).)
9. If x is an orthogonal patch in a geometric surface M, show that
(intrinsic derivatives). (Hint: For the associated frame field; compute
/ \ ( Xv \ Zb* x "
«*(*.) = Vv^A V^ " Xvu Weg
Then using the formula for w n given in Chapter VI, Section 6, show that
326 RIEMANNIAN GEOMETRY [Chap. VII
W21 ( xtt )=__ = xttv __.
Find an easier proof in the special case where M is a surface in E 3 .
10. (Continuation) Show that y uv = y vu for an arbitrary patch y. (Hint
There exists an orthogonal patch x such that x = y(u,v).)
11. If there exists a nonvanishing vector field W on M such that V V W =
for all V, show that M is flat. Find such a vector field on a cylinder
in E 3 .
4 Geodesies
Geodesies in an arbitrary geometric surface generalize straight lines in
Euclidean geometry. We have seen that a straight line a(t) = p + tq is
characterized infinitesimally by vanishing of acceleration; thus
4.1 Definition A curve a in a geometric surface M is a geodesic of M
provided its acceleration is zero; / = 0.
In other words, the velocity a of a geodesic is parallel: geodesies never
turn. Recall that a parallel implies \\a  constant; so geodesies have
constant speed.
Because acceleration is preserved by isometries (Exercise 7 of Section
3), it follows that geodesies are isometric invariants. (A direct proof
appears in Exercise 1 of Section 5, Chapter VI). In fact, if F: M — > N is
merely a local isometry, then F carries each geodesic a of M to a geodesic
F(a) of N — for F is locally an isometry, as discussed in Chapter VI, Sec
tion 4.
The general definition of geodesies given above agrees with that of
5.7 in Chapter V when M is a surface in E 3 , for we can deduce from Lemma
3.4 that the intrinsic acceleration of a curve ainMcE is the component
tangent to M of its Euclidean acceleration. Thus the former is zero if
and only if the latter is normal to M.
Suppose that a: I — » M is a curve in an arbitrary geometric surface M
and Ei, E 2 is a frame field on M . Throughout this section we use the nota
tion
a = ViEi + v 2 E<t and a" = A^i + A 2E2
for the velocity and acceleration of a. From Section 3 we know that these
components of acceleration are
A x = Vi + V 2 a)2i(ot')
A 2 = v 2 ' + Viwn(a)
Sec. 4] GEODESICS 327
and are realvalued function on the interval I. Our main criterion for a
to be geodesic is thus A x = A 2 = 0. Using orthogonal coordinates we now
rewrite these equations in a more informative way.
4.2 Theorem Let x be an orthogonal coordinate patch in a geometric
surface M. A curve a{t) = xfait), a*(t)) is & geodesic of M if and only if
ai " + hi {Euai ' 2 + 2E ^ a * " aa2 ' 2} = °
a 2 " + ~ \E v a x 2 + 2(? M a 1 V f G v a 2 2 } = 0.
We shall subsequently refer to these equations as A x = and A 2 = 0.
Note that they are symmetric, in the sense that the reversals 1 <> 2, u <> v,
E <> G turn each one into the other. In this context we shall always under
stand that the functions E, G and their partial derivatives E u , E v , • • • are
evaluated on (a x , o 2 ), and hence become functions on the domain J of a.
Proof. The velocity of a is a = o/x u + a 2 x v , so in terms of the asso
ciated frame field of x (Chapter VI, Section 6), we have
«' = (a x VE)E X + (a 2 VG)E 2 .
Thus the acceleration components A x , A 2 defined above become
A x = (a x VE)' + (a 2 ' VG)<*i(a)
A 2 = (a 2 ' \/G)' + (a x ' VE)o>n(d).
Using the formula for a>i 2 from Chapter VI, Section 6, we find
/ '\ t ' i ' ^ ('vE)v „ i , ('vG)u i / 9 \
o>n{<* ) = wi 2 (aix„ + a 2 xj = ^= ai + — 7= a 2 U;
When this is substituted in (1), the geodesic equations A x = A 2 =
become
(a/ VEY + (VEWa.' " V ^ ( ^ )m a/ 2 =
(1)
(a,' VG)'  ^V^ «i' 2 + (VG)4h'<h = 0.
(3)
Standard calculus computations will transform (3) to the form stated in
the theorem. We merely remind the reader that in a Leibnizian expansion
such as
77"
(a x VE)' = «i" VE + a x J^= ,
328 RIEMANNIAN GEOMETRY [Chap. VII
the notation E is short for E(a 1} (h), so
E' = E u ai + E v a 2 . 
4.3 Theorem Given a tangent vector v to I at a point p, there is a
unique geodesic a of M such that
a(0) = p, a'(0) = v.
Thus there are lots of geodesies in any geometric surface, and each is
completely determined by its initial position and velocity. In E 2 for ex
ample, the geodesic determined by v at p is the straight line a(t) = p + t\.
Proof. Let x be an orthogonal patch in M with p = x(u , v ), and write
v = cox u + d x v . The geodesic equations in Theorem 4.2 have the form
a/' = /i (ai, a 2 , ai , a 2 )
(h = j2 (ai, «2, ai , «2 )
Furthermore, a will satisfy the given initial conditions if and only if
ai(0) = wo a/(0) = Co
a 2 (0) = vo a 2 ' (0) = d .
Now the fundamental existence and uniqueness theorem of differential
equations asserts that there is an interval / about on which are defined
unique functions a u a 2 which satisfy (1) and (2). Thus a = x(ai, a 2 ) is
the only geodesic defined on / such that a(0) = p, c/(0) = v. 
This proof is not entirely satisfactory, because the interval / may be
unnecessarily small. We describe briefly a way to make it as large as possi
ble. Suppose that ct\: Ii — > M and a 2 : h —> M are geodesies satisfying the
same initial conditions at t = 0. Using the uniqueness property above, we
can deduce that ai = a 2 on the common part of I x and 7 2 . Applying this
consistency result to all such geodesies, we obtain a single maximal geodesic
a: I — * M satisfying the initial conditions. (The interval / is the largest
possible.) Intuitively this means we simply let the geodesic run as far as
it can.
4.4 Definition A geometric surface M is geodesically complete provided
every maximal geodesic is defined on the whole real line R.
Briefly: geodesies run forever. A constant curve is trivially a geodesic,
but excluding this case, every geodesic has constant nonzero speed. Thus
geodesic completeness means that all nontrivial (maximal) geodesies are
infinitely long — in both directions. For example, E 2 is certainly complete,
and the explicit computations in Example 5.8 of Chapter V show that
Sec. 4] GEODESICS 329
N = J(T)
FIG. 7.8
spheres and cylinders in E 3 are. More generally, all compact geometric
surfaces are complete, as are all surfaces in E 3 of the form M: g = c (con
sequences of Theorem 15, Chapter 10, of Hicks [5]). Removal of even a
single point from a complete surface will destroy the property since geo
desies formerly passing through the point will be obliged to stop.
There is a Frenet theory of curves in a geometric surface M which
generalizes that for curves in the plane (Exercise 8 of Section 3 in Chapter
II). Because M has only two dimensions, torsion cannot *oe defined. If M
is oriented, however, curvature can be given a geometrically meaningful
sign, as follows. If 0: / — > M is a unitspeed curve in an oriented geometric
surface, then T = fi' is the unit tangent vector field of 0. To get the principal
normal vector field N, we "rotate T through +90°," denning
N = J(T),
where J is the rotation operator from Exercise 5, Section 1 (Fig. 7.8).
Then the geodesic curvature k„ of is the realvalued function on / for which
the Frenet formula T' = k N holds. Thus k is not restricted to nonnegative
values as in the case of curves in E 3 : k > means that T — hence 0— is
turning in the positive direction as given by the orientation of M, and
Kg < means negative turning.
4.5 Lemma Let be a unitspeed curve in a region oriented by a frame
field E u Ei. If <p is an angle function from Ei to #' along /3, then
Kg = j — h 0>l 2 (j3 ).
ds
Proof. By definition of angle function (Exercise 15 in Chapter VI, Sec
tion 7), we have
T = 0' = cos <p Ei + sin <p E 2 .
Since the orientation derives from this frame field,
J(E X ) = E 2 and J (E 2 ) = E lt
so
330 RIEMANNIAN GEOMETRY [Chap. VII
N = J(T) = sin <pEi + cos v E 2 .
Using the derivative formula on page 320, we find
T = /3" = { — sin tp <p' + sin <p wsiOs')}^
+ {cos <p <p f + cos ^ wbOs')}^.
But K g = T'N = T'J(T), so using the formulas for T' and /(T 7 ), we get
K g = (cosV + sinV)0' + coi 2 (/3')) = */ + wiaOS 7 ). 
For example, in E 2 the natural frame field has o>i 2 = 0, and <p becomes
the usual slope angle of the curve 0. Thus the result reduces to k = a\p/ds,
which in elementary calculus is often taken as the definition of curvature.
For an arbitraryspeed regular curve a in M, the Frenet apparatus
T, N, K g is defined — just as in Chapter II, Section 4 — by reparametrization.
Furthermore the same proof as for Lemma 4.2 of Chapter II shows
dv
di
a = vT a." =%T + Kg v 2 N (*)
where v = \\ a \\ is the speed function of a.
4.6 Lemma A regular curve a in M is a geodesic if and only if a has
constant speed and geodesic curvature k = 0.
Proof. Since v > 0, we have a" = if and only if (dv/dt) = k = 0. 
/
The equations ( ) also show that a has geodesic curvature zero if and only
if a and a" are always collinear. Such curves are sometimes called geo
desies: to get a geodesic in the strict sense of Definition 4.1 it suffices to
reparametrize a to a constant speed curve. (Proof: k is unaffected by
reparametrization.) In contexts where parametrization is of some im
portance we shall call a curve with K s = 0a pregeodesic.
Computation of explicit formulas for the geodesies of a given geometric
surface is rarely a simple task. Our purpose, however, is not to collect
formulas, but to study the general behavior of geodesies. Before continuing
this study we are going to examine an important special case in which a
considerable amount of concrete information about geodesies can often
be obtained with only a minimum of computation.
4.7 Definition A Clair ut parametrization x: D —*■ M is an orthogonal
parametrization for which E and Cr depend only on u, that is, F = and
E v = G v = 0.
For example, the usual parametrization of a surface of revolution is a
Clairut parametrization.
4.8 Lemma If x is a Clairut parametrization, then
Sec. 4] GEODESICS 331
1. All the nparameter curves of x are pregeodesics, and
2. A ^parameter curve, u = u , is a geodesic if and only if G„(w ) = 0.
Proof. For (1) it suffices by a remark above to show that x u and x uu
are collinear. Since x„ and x„ are orthogonal, this is equivalent to
X V 'X UU = U.
But the following equations imply this result:
= tLv = (.Xii^XtiJu == ZX U *X UV
= fr u = \x u *x v ) u == x uu *x v ~r x u *x vu .
Similarly, for (2), the vparameter curve, u = u , is pregeodesic if and
only if x vv (u , v)'x u (u , v) = 0. The following equations show that this is
the case if and only if G u (uq) = 0.
U = f j = Xtiv * X v ~j~ X u • X vv
G u (uo) = G u (u , v) = 2x vu (u , v)»x v (uo, v) for all v.
(Recall that x uv = x vu .) We have not used the condition G v = 0; its only
effect is to show that yparameter pregeodesics are in fact geodesies, since
it means that vparameter curves have constant speed. I
In the case of a surface of revolution this lemma provides an intrinsic
proof that meridians are geodesies and that a parallel, u = u , is geodesic
if and only if ti (u ) — 0. (See Exercise 3 of Chapter V, Section 5).
Because of the preceding lemma we shall think of a Clairut parametriza
tion as a "flow" whose streamlines are its wparameter geodesies, and we
shall measure the behavior of arbitrary geodesies relative to this flow.
4.9 Lemma If a = x(a u a^) is a unit speed geodesic, and x is a Clairut
parametrization, then the function
c = G(ai)ch = s/G(ai) sin <p
is constant, where ip is the angle from x„ to a . Hence a cannot leave the
region for which G ^ c .
We call the constant c thus associated with each geodesic a the slant
of a, since^ — in combination with G — it determines the angle <p at which a
is cutting across the wparameter streamlines of x (see Fig. 7.9).
Proof. Because E v = G v = for a Clairut parametrization, the equation
A 2 = of Theorem 4.2 reduces to
a 2 + y7 ai a 2 =0
332
RIEMANNIAN GEOMETRY
[Chap. VII
FIG. 7.9
But this is equivalent to the constancy of c = G(h, since
(Gat')' = G' a* + G a 2 " = GUi' a,' + G a 2 "
To show that c = y/G sin <p, compare the two equations
a 'X v = (ai x u + a 2 x„)»x r = (ja 2 ' = c
a'x, =  a   x„  cos (f "~ ^) = V& sin <p
It follows immediately from  sin <p  ^ 1 that 6r ^ c 2 .
I
If a is moving in a direction in which G is increasing, then the constancy
of c = y/G sin <p shows that ^> is decreasing: a is forced to turn more in the
direction of the flow. On the other hand, if G is decreasing along a, then
a cuts across the wparameter geodesies at everincreasing angles. Some
interesting consequences are presented in Exercises 11 and 12. This inter
pretation is particularly simple on a surface of revolution (Exercise 13).
We can add to the lemma above the equation
ai = =fc
\/G  i
Veg
(1)
In fact, since a has unit speed we have 1 = «'•«' = Zfa/ 2 + Ga 2 ' 2 . In this
equation substitute
a 2 =
G
(2)
(from Lemma 4.9), and solve for a/ to obtain equation (1).
Conversely, a straightforward computation shows that if a/ is nonzero,
the equations (1) and (2) imply that a = x(ai, a*) is a unit speed geo
Sec. 4] GEODESICS 333
desic. Furthermore, a/ nonzero is a necessary and sufficient condition for an
arbitrary curve a to have a reparametrization of the form
j8(w) = x(w, v(u)).
The point of all this is that we can now give a relatively simple criterion
for a curve of this form to be a pregeodesic — and thereby determine the
routes of the geodesies in a region with a Clairut parametrization. (The
special parametrization of /3 misses essentially only the geodesies given by
(2) of Lemma 4.8; see Exercise 12.)
4.10 Theorem A curve /3(w) = x(u, v(u)), where x is a Clairut para
metrization, is a pregeodesic if and only if
— = ±c\/E
d^ ~ VGVG  c 2
The constant c is then the slant of /3.
Proof. The argument is simply an exercise in change of parametrization.
Let a be a unit speed reparametrization of p derived just as in Chapter II
from an arc length function s for 0. Thus is pregeodesic if and only if
a is geodesic. Let a,\ be the inverse function of s (so Oi does not vanish).
Then
a = (8(ai) = x(ai, y(«i)),
and we set a% = v{a\). The remarks above show that a is a geodesic (of
slant c) if and only if
a\ = — _T c , a 2 ' = ^ (E,G evaluated onci) (1)
y/EG "
If these equations hold, then by elementary calculus
dv _ azis) _ ±c \/E
du ~ ax'(s) ~ y/Q VG  c 2
(2)
where the substitution of s (inverse function of ai) makes E and G merely
functions of u. Conversely, if (2) holds, we deduce (1) using the equation
Ea x ' 2 + Go* 2 = 1 which expresses the fact that a has unit speed. 
Since the formula above for dv/du depends only on u, by the fundamental
theorem of calculus it can be written in integral form as
v(u) = v(uo) ± / , nW7T 
'o Vg Vg  c 2
Thus for a Clairut parametrization — in particular, for a surface of revo
334 RIEMANNIAN GEOMETRY [Chap. VII
lution — the computation of pregeodesics is reduced to a single integration.
This is, of course, a vastly simpler criterion than the secondorder differ
ential equations i)f Theorem 4.2. Unfortunately, however, the integration
can rarely be carried out in terms of elementary functions.
4.11 Example Routes of Geodesies
(1) The Euclidean plane E . We begin with a surface whose geodesies we
already know, but to illustrate the preceding result we shall find their
routes in terms of the polar parametrization
x(u, v) = (u cos v, u sin v).
Since E — 1, F = 0, and G = w 2 , this is a Clairut parametrization. The
wparameter geodesies are just the radial lines through the origin. All
others may be parametrized as /5(w) = x(u, v(u)), where by Theorem 4.10
dv_ = ±c ^ d / _ x c\
du u \Zu 2 — c 2 du\ u)
Hence v — v = ± cos 1 (c/w), or u cos (v — v ) = c, which is the polar
equation of a straight line. The slant c has geometrical significance as the
distance from the line to the origin.
(2) The hyperbolic plane H. Polar coordinates are a more natural choice
in this case since the function g giving the geometric structure of H de
pends only on distance to the origin, f Thus if
x(u, v) = (u cos v, u sin v), < u < 2,
thengf(x) = 1 — w 2 /4. (We write simply g henceforth.) Now
1 u
E = x„ ° x„ = — , F = 0, G = x„ o x „ = — ;
r g 2
thus x is a Clairut parametrization. By Lemma 4.8 the wparameter curves
— Euclidean lines through the origin — are routes of geodesies of H. By
Theorem 4.10, /3(w) = x(w, v (u)) is pregeodesic provided
<fo _ ±(cg/u 2 )
du Vl  (cg/u) 2 ' { )
To carry out the required integration, set
w = \1 \ — }, where a =
u\ 4/
Vl + c 2 "
Then a straightforward computation yields
t An even better choice, the hyperbolic polar coordinates discussed in Example
5.5, gives rise to the substitution used later in this example.
Sec. 4
GEODESICS
335
dv
du
T dw/du
Vl  w 2
(2)
Hence
Thus
v — v = ±cos w, or cos (v — Vo) = w = 
4w
SO + s)
u + 4 cos (v — Vo) =
a
(3)
Using the law of cosines in a diagram similar to that in Fig. 7.10, one finds
the polar equation of a circle of radius r, centered at x(u , v ).
2 , 2
U + Uq
2u u cos (y — v ) = r 2 .
(4)
Comparison with equation (3 ) shows that the route C of /3 is a Euclidean
circle with wo 2 — r 2 = 4. Since w > 2, the center of C lies outside the
hyperbolic plane H: x 2 + y < 4. One can see from Fig. 7.10 that the circle
C is orthogonal to the rim x 2 + y 2 = 4 of #. Of course, lies in the open
arc of C inside H, and we deduce from Theorem 4.2 that fills this arc.
Conc/us/on. The routes of the geodesies of the hyperbolic plane H are
the portions in H of : all Euclidean straight lines through the origin, and all
Euclidean circles orthogonal to the rim of H.
The argument in Example 2.4 suggests that the geodesies of H have
infinite length (a formal proof is given in Exercise 1 of Section 5); thus
H is geodesically complete.
The geodesies of the hyperbolic plane bear comparison with those of
the Euclidean plane. Around 300 B.C. Euclid established a celebrated
set of axioms for the straight lines of his plane. The goal was to derive its
x(w , t'o)
FIG. 7.10
336 RIEMANNIAN GEOMETRY [Chap. VII
FIG. 7.11
geometry from axioms so overwhelmingly reasonable as to be "selfevi
dent." The most famous of these is equivalent to the 'parallel postulate:
If p is a point not on a line a, then there is a unique line through p which
does not meet a. Over the centuries this postulate began to seem some
what less selfevident than the others. For example, the axiom that two
points determine a unique straight line might be checked by laying down
a (perhaps long, but still finite) straight edge touching both points. But
for the parallel postulate one would have to travel the whole infinite length
of j8 to be sure it never touches a. Thus tremendous efforts were expended
in trying to deduce the parallel postulate from the other axioms. The
hyperbolic plane H offers the most convincing proof that this cannot be
done. For if we replace "line" by "route of a geodesic," then every Euclid
ean axiom holds in H except the parallel postulate. For example, given
any two points it is easy to see that one and only one geodesic route runs
through them. But it is clear from Fig. 7.11 that in H there are always an
infinite number of geodesic routes through p that do not meet a. When the
implications of this discovery were worked out, what was destroyed was
not merely the modest hope of deducing the parallel postulate, but the
whole idea that E 2 is, in some philosophical sense, an Absolute, whose
properties are "selfevident." It had become but one geometric surface
among the infinitely many others discovered by Riemann.
EXERCISES
1. Show that a reparametrization a(h) of a nonconstant geodesic a is
again a geodesic if and only if h has the form h(t) = at + b.
Sec. 4] GEODESICS 337
FIG. 7.12
2. Denote by y v the unique geodesic in M with initial velocity v. For
any number a, show that y av (t) = y v (at) for all t.
3. Let V be a vector field on a geodesic a. Show that V is parallel if and
only if  V  is constant and V makes a constant angle with a .
4. In the sphere 2, let nbe the north pole, pi and p2 points on the equator.
Consider the broken curve /3 following a meridian from n to p x , the
equator from pi to p2, then a meridian from p2 back to n. Prove that
the holonomy angle of /3 is the angle at n between the two meridians.
5. Find the routes of the geodesies in the stereographic sphere, (1) of
Example 2.3.
6. In the Poincare" half plane, show that the routes of the geodesies are:
all semicircles with centers on the waxis, and all vertical lines. (Hint:
x (w, v) = (u, v) is a Clairut patch "relative to v," so in the text
equations, reverse u and v, and E and G.) See Fig. 7.12.
7. Let a be a unit speed curve such that a is never collinear with Ei,
where E u Ei is a frame field. If a" = AiEi + A 2 E 2 , show that the
single equation A\ = implies that a is a geodesic.
8. In the projective plane of radius r (Exercise 6 of Section 2), prove:
(a) The geodesies are simple closed curves of length irr.
(b) There is a unique geodesic route through any two distinct points.
(c) Two distinct geodesic routes meet in exactly one point.
(Hint: every geodesic in 2 is the image under the projection P:
2 — > S of a geodesic in the sphere 2.)
9. If a is a curve in M with speed v > 0, prove:
(a) The geodesic curvature n g of a is a" *J (a )/v . Hence if M is a
surface in E 3 , k = Ua X a." /v.
(b) If a has unit speed, then in Ex. 7 of Chapter V, Section 5, the
vector field V is the unit normal N of a, and the function g is
geodesic curvature k .
10. Let M be the plane with the origin deleted, and furnish M with the
conformal geometric structure for which g = r = \/u 2 f v 2 .
Find the Gaussian curvature of M and the routes of its geo
desies. Show that M is isometric to a surface in E .
11. Let x be a Clairut parametrization, and let a = x(ai, a^) be a unit
338
RIEMANNIAN GEOMETRY
U = tt,
[Chap. VII
wparameter curves
FIG. 7.13
speed geodesic with slant c. Suppose that a starts at the point
a(0) = x(«i(0), a 2 (0)) = x(u ,v )
and that ai (0) > 0. If there is a number u > u such that G(u) =
c, let Wi be the smallest such number. Then the vparameter curve
0(v) = x(wi, v)
is called a barrier curve for a. Prove that:
(a) a comes arbitrarily close to /3
(b) If j8 is a geodesic, a does not meet /S (thus a asymptotically ap
proaches/8).
12. (Continuation) If the barrier curve /3 is not a geodesic, it can be shown
thatadoesmeetjS.flf a (£*) is the meeting point, show that cti (t*) =
and that a± changes sign at t*. Thus a bounces off fi as shown in Fig.
7.13. (Hint: prove that af (t*) < 0.)
13. Let a be a geodesic on a surface of revolution.
(a) Show that the slant of a is c = h sin <p, where h(t) is the distance
from a(t) to the axis of revolution, and <p gives the angles at
which a cuts the meridians of M .
(b) Deduce that a cannot cross a parallel of radius \c\.
1 4. Let a be a geodesic of slant c on the paraboloid of revolution
M: z = x + y 2 
Find the minimum value of z(a), that is, the lowest height to which
a descends. (Hint: Use a Monge patch.)
15. Prove that no geodesic on the bugle surface (V.6.6) can be defined
on the whole real line.
16. On a torus of revolution, let a be a geodesic that at some point is
t We assume, of course, that a remains in the region parametrized by x.
Sec. 5] LENGTHMINIMIZING PROPERTIES OF GEODESICS 339
tangent to the top circle (u = ir/2). Show that a remains always on
the outer half of the torus (— t/2 ^ u S t/2) and travels around
the torus oscillating between the top circle and bottom circle.
17. Let C be a catenoid (Example 6.1 of Chapter V) with c = 1, and let
a be the geodesic such that
a(0) = x(wo, Vo), Wo 5^ 0,
and a (0) makes angle <p with the meridians. (Note that <p and ir — <po
determine different parametrizations of the same geodesic.) For which
values of <p does a cross the minimal circle, u = 0, of C?
18. A Liouville parametrization x: D — » ilf is an orthogonal parametriza
tion for which E = G = U + V, where £7 is a function of u only and
7 a function of v only. If a — x(a u ch) is a unit speed geodesic ex
pressed in terms of such a parametrization, show that
Uish) sin 2 (p — V((h) cos 2 <p
is a constant, where <p is the angle from x u to a .
19. Let Ei, E 2 be a frame field on a geometric surface M. For i = 1, 2>
let *»(p) be the geodesic curvature at p of the integral curve of E<
through p.
(a) Prove that K = Ei[k 2 ]  E 2 [ki]  k x 2  k 2 \
(b) Test this formula on an arbitrary surface of revolution, using
the frame field in Example 6.4 of Chapter VI.
(Hint: for (a), prove coi 2 (i?i) = k»)
5 LengthMinimizing Properties of Geodesies
The previous section considered geodesies as straightest curves; now we
investigate their character as shortest curves. The basic problem is, roughly
speaking, to find the shortest route from one point to another in a geo
metric surface. For E 2 the solution is simple: Given any two points p and
q, there is a unique straightline segment from p to q, and this is shorter
than any other curve from p to q (Exercise 11 of Section 2 of Chapter II).
For an arbitrary geometric surface M, the situation is more interesting.
In the first place, there may be no shortest curve from p to q (Exercise 3
of Section 4 of Chapter VI). And even if there is one, it may not be unique.
For example, we shall soon prove the expected result that on a sphere, all
semicircles from the north pole to the south pole have the same shortest
length. To make the terminology precise, we use the notion of intrinsic
distance (Chapter VI, Section 4).
340
RIEMANNIAN GEOMETRY
[Chap. VII
5.1 Definition Let a be a curve segment from p to q in M . Then
(1) a is a shortest curve segment from p to q provided L(a) = p(p, q)
(2) a is the shortest curve segment from p to q provided that
L(a) = p(p, q)
and that any other shortest segment from p to q is merely a reparametriza
tion of a.
In the first case we shall also say that a minimizes arc length from p to q;
the definition means that if fi is any other curve segment from p to q, then
L(fi) * L(a).
In the second case, we say that a uniquely minimizes arc length. "Unique
ness" must be interpreted liberally enough to allow for reparametrization,
since monotone reparametrization (Exercise 10 of Section 2 of Chapter II )
does not change arc length.
All such shortest curves will turn out to be geodesies (Lemma 5.8).
Our first main result (Theorem 5.6) will show that short enough geodesic
segments behave as well in an arbitrary geometric surface as they do in E 2 .
Some preparatory work is needed first.
In the Euclidean plane, if one is interested in the distance to the origin,
it is natural to use polar coordinates, for then the distance from to
x(u, v) = (u cos v, u sin v)
is simply u. We shall now generalize this parametrization to the case of an
arbitrary geometric surface M . As for E 2 , the wparameter curves will be
geodesies radiating out from some fixed point p of M. Such geodesies may
conveniently be described as follows: If v is a unit tangent vector at p,
let 7 V be the unique geodesic which starts at p with initial velocity v. Now
we assemble all these geodesies into a single mapping as follows:
5.2 Definition Let e x , e 2 be a frame at the point p of M . Then
x(u, v) = 7,
cos tie, + sin
•e,(w)
FIG. 7.14
is the geodesic polar mapping of M
with pole p.
Here the domain of x is the larg
est region of E on which the for
mula makes sense. A choice of v fixes
a unit tangent vector
v = cos vei + sin vez
at p (Fig. 7.14). Then the wparam
Sec. 5]
LENGTHMINIMIZING PROPERTIES OF GEODESICS
341
eter curve
x(w, v) = y r (u)
is the radial geodesic with initial velocity v. Since  v  = 1, this geodesic
has unit speed, so that the length of y v from p = 7 V (0) to y y (u) is just u.
In the special case where ei, *% is the natural frame at the origin of E 2 ,
the geodesic polar mapping becomes
x (u, v) = 7 cos „ ei + sin VBi (u)
= + w(cos vei + sin Vd)
= (u cos v, u sin v).
Thus x is a generalization of polar coordinates in the plane.
The pole p is a trouble spot for a geodesic polar mapping. To clarify the
situation near p, we define (in the situation described in Definition 5.2)
a new mapping
y(u,v) = Tuei +»e 2 (l)
Differential equations theory shows that y is differentiable, and it is easy
to check that y is regular at the origin. Thus by the inverse function the
orem, y is a diffeomorphism of some disc D t : u 2 + v < i onto a neighbor
hood dl f of p. We call 9l £ a normal neighborhood of p. In the special case
M = E 2 , y is just the identity map y(u,v) = (u,v),so for arbitrary M, y
is a generalization of the natural (rectangular) coordinates of E 2 .
5.3 Lemma For a sufficiently small number e > 0, let S t be the strip
< u < e in E 2 . Then a geodesic polar mapping x: S t — > M with pole p
parametrizes a normal neighborhood 3l t of p — omitting p itself, (see Fig.
7.15).
—s t 
FIG. 7.15
342 RIEMANNIAN GEOMETRY [Chap. VII
FIG. 7.16
Proof. Note that x bears to y the usual relationship of polar to rectan
gular coordinates; that is,
X (U, V ) = y C os vei + sin VB2 V^/ = Tu cos iei + m sin »e2 \ /
= y(w cos v, u sin v)
where we use the identity y v (u) = 7« v (l) from Exercise 2 of Section 4.
Now this formula expresses x as the composition of two regular mappings:
(1) The Euclidean polar mapping (u, v) — > (u cos v, u sin v) which
wraps the strip S t around the disc D t , and
(2) The onetoone mapping y of D t onto 9l e .
Thus x is regular and carries S t in usual polarcoordinate fashion onto the
neighborhood 9l £ — omitting only the pole. 
We draw a fundamental consequence: If q = x(wo, Vo) is any point in a
normal neighborhood 91 of p, then there is only one unit speed geodesic from
p to q which lies entirely in 91, namely, the radial geodesic
y(u) = x(u, v ) ^ u ^ uq.
(Proof: Any unitspeed geodesic starting at p is, by the uniqueness of
geodesies, a wparameter curve of the polar parametrization. As suggested
by Fig. 7.16, all except v = v + 2im lead out of 91 without hitting q, and
different choices of n still give the same geodesic y by the usual ambiguity
of polar coordinates).
5.4 Lemma For a polar geodesic parametrization, E = 1, F = 0, G > 0.
Proof. Since the wparameter curves are unitspeed geodesies, we have
E = x M »x M = 1, and x uu = 0.
Thus
P u == \X U * Ti. v ) u X u * X. vu jl u * JL UV ^C'u O,
so F is constant on each wparameter curve. Now the functions E, F, G
are welldefined even when x is not restricted to the strip S f . The vparam
eter curve v — > x(0, v) is simply the constant curve at the pole p, so
Sec. 5] LENGTHMINIMIZING PROPERTIES OF GEODESICS 343
x v (0,v) = 0.
But then F(0, v) = for all v, and since F u = 0, we conclude that F is
identically zero. Because x (now restricted once more to the strip S f ) is a
parametrization, that is a regular map, we know that EG — F 2 = EG is
never zero. Hence G > 0. 
5.5 Example We explicitly work out geodesic polar parametrizations
in two classic cases.
(1) The unit sphere 2 in E . For simplicity let p be the north pole (0, 0, 1 ).
To get the geodesies radiating out from p as in Fig. 7.17, we change the
geographical parametrization to
x(u, v) = (sin u cos v, sin u sin v, cos u).
Each wparameter curve is indeed a unitspeed parametrization of a great
circle, hence is geodesic. For u = we find
x M (0, v) = (cos v, sin v, 0)
= cos vei + sin ve 2
where
ei = C/i(p), e 2 = t/ 2 (p).
Thus by the uniqueness of geodesies,
X \U, V ) = 'Ycos vei + sin v ei \M )
which shows that x as denned above is the polar geodesic mapping (Defi
nition 5.2) . It is easy to see that the largest possible normal neighborhood
9l e of p occurs when e = ir, for on the strip &, x is a polar parametrization
of all the sphere except the north and south poles.
(2) The hyperbolic plane H (Example 2.4). We choose p = (0, 0) and
FIG. 7.17
344
RIEMANNIAN GEOMETRY
[Chap. VII
«i = Ui(p), 62 = £Mp) (Since the function g is 1 at the origin, this is a
frame.) We know from Example 4.11 that the geodesies of H through the
origin follow Euclidean straight lines. Thus for any number v, the curve
a{t) = (t cos v, t sin v)
is at least a pregeodesic of the type we are looking for. In Example 2.4 we
found the arc length function s(t) = 2 tanh _1 (</2) for such a curve; thus
s — > a I 2 tanh  1 = ( 2 tanh  cos v, 2 tanh  sin v 1
is the unitspeed reparametrization of a. Shifting notation from s to u, we
obtain
x(w, v) = I 2 tanh  cos v, 2 tanh  sin
•)
2 ' 2
Since the wparameter curves of x are unitspeed geodesies and
x„(0, v) = cos vei + sin ve%,
we conclude as in (1) that x is a geodesic polar mapping. The normal
neighborhood in this case is the entire surface H.
5.6 Theorem For each point q of a normal neighborhood 9l £ of p the
radial geodesic segment in 9l t from p to q uniquely minimizes arc length.
Proof. Let x be the polar parametrization of the normal neighborhood
?fl f . If
q = x(wo, v ),
the radial geodesic segment is
y(u) = x(u, v ), ^ u ^ Wo
Now let a be an arbitrary curve segment from p to q in M ; we may arrange
for a to be denned on the same interval as y.
We begin by proving
L(y) ^ L(a).
(1)
FIG. 7.18
First consider the case (Fig. 7.18) where
a stays in the neighborhood 3l e . We may
assume that once having left p, a never
returns to p — if it did, throwing away
this loop would only shorten a. Thus it
is possible to write
a(t) = xfa (0,02(0).
Sec. 5] LENGTHMINIMIZING PROPERTIES OF GEODESICS 345
Since a(0) = p and a(u ) = q, we have
ai(0) = ai(wo) = u
(h(0) = v a 2 (w ) = v + 2im.
(The term 2m results again from the nonuniqueness of angles in polar
coordinates. )
Since for x we have E = 1 and F = 0, the speed of a is
II a ' II = vV 2 + Ga 2 ' 2 .
Now
Hence
Va/ 2 + Ga 2 ' 2 ^ \/V 2 =  ax  ^ a/ (3)
<ft
(4)
L(a) = [ ° Voi' f + GV 2 d« ^ / a/
Jo J o
= ax(uo) — ai(0) = Wo
where the last step uses (2). But the radial geodesic y has unit speed, so
/•«0
L(y) = I dt = wo,
Jo
and we conclude that
L(y) ^ L{a).
If a does not stay in 9l e we have strict inequality, L{y) < L(a). For a
must cross the poZar circle, u = w — indicated by a dashed line in Fig. 7.18
— to escape from 9l«.f But by the proof above it must already have arc
length at least u = L(y) when it reaches the circle.
Now we prove the uniqueness assertion:
If L(a) = L(y), then a is a reparametrization of 7 (5)
The argument above shows that if L(a) = L(y), then a stays inside
3fl« and the inequality in (4) becomes equality. The latter implies that
V«i' 2 + Ga 2 ' 2 = th. Since G > we conclude from (3) that
ax ^ 0, <U = 0. (6)
Thus Oi has constant value v (so n = in (2 ) ) and
a(t) = x(ai(0, v ) = y(a r (t)),
showing that a is in fact a monotone reparametrization of 7. 
t A careful proof will involve the Hausdorff axiom (Ex. 5 of IV.8) which we
assume throughout this chapter.
346
RIEMANNIAN GEOMETRY
[Chap. VII
FIG. 7.19
This fundamental result shows, as we remarked earlier, that if points
p and q are close enough together, then — as in Euclidean space for arbitrary
points — there is a unique geodesic segment from p to q which is shorter
than any curve from p to q. (Unlike the Euclidean case however, there
may be many other nowshortest geodesies from p to q.) If x is a geodesic
polar parametrization at p we shall also call the route C ( of the vparameter
curve, u = e, the polar circle of radius e at p. (Fig. 7.19). Theorem 5.6
shows that C t does in fact consist of all points at distance e from p.
In special cases where large normal neighborhoods are available, this
local information may be decisive.
5.7 Example Lengthminimizing properties of geodesies on the sphere 2
of radius r. By a mere change of scale we can conclude from Example 5.5
that each point p of 2 has normal neighborhood 9l, r : all of 2 except the
point, — p, antipodal to the pole p. Hence Theorem 5.6 implies:
(a) If two points p and q of 2 are not antipodal (that is, q 5* — p),
then there is a unique shortest curve y from p to q. But we know all the
geodesies of 2: 7 can only be the one that folllows the shorter arc of the
great circle through p and q.
(b) Intrinsic distance p on 2 is given by the formula
p(p, q) = rd
where t? (0 ^ t? ^ x) is the angle from p to q in
E (Fig. 7.20). If p and q are not antipodal, this
follows from (a), since
P(p>q) = L(y) = re.
As q moves toward the antipodal point — p of p
we deduce by continuity that p(p, — p) = rir.
Hence
fig. 7.20
Sec. 5] LENGTHMINIMIZING PROPERTIES OF GEODESICS 347
FIG. 7 21
(c) There are infinitely many minimizing geodesies from a point p on
2 to the antipodal point — p, namely (constantspeed parametrizations of)
semicircles from p to —p. (Proof: These all have length rir = p(p, — p).)
(d) No geodesic segment y of length L(y) > vr can minimize arc length
between its end points. This follows immediately from the fact that in
trinsic distance p never exceeds ur. It is geometrically clear, since if y
starts at p, its length exceeds irr as soon as 7 passes the antipodal point
—p. But then the other arc 7* of the same great circle is shorter than 7.
Suppose that a is a curve segment in M from p to q, and /3 is a curve
segment from q to r. Now a and /3 cannot generally be united to form a
single (differentiable) curve from p to r, since there may be a "corner"
at q as in Fig. 7.21. Using the techniques of advanced calculus, one can
"round off" this corner, obtaining a curve segment 7 from p to r which
(to state the weakest theorem) is only slightly longer than a and /3. Ex
plicitly, for each e > there is a 7 such that L(y) ^ L(a) + L(0) + e.
It follows that intrinsic distance satisfies the triangle inequality. In fact,
given points p, q, and r the definition of intrinsic distance shows that for
any e > there exist curves a and /3 as above such that
LCIC CA1SL UU1VCS « £UU1 jJ na »UUVC OUl^ll uuau
L(a) ^ p(p, q) + e, L(/3) ^ p(q, r) +
e.
Rounding off the corner at q costs at most another e: We get a curve
segment 7 from p to r such that
p(p, r) ^ L(y) ^ p(p, q) + p(q, r) + 3e.
But since e is arbitrary, we conclude that
p(p, r) ^ p(p, q) + p(q, r).
5.8 Lemma If a is a shortest curve segment in M from p to q, then a
is geodesic.
Proof. We shall prove that if a : [a, b] — > M is a curve segment from p to
q which is not a geodesic, then L(a) > p(p, q). But if a is not geodesic,
then at some time £ the acceleration a" (to) is not zero. By continuity a."
is nonzero near t , so we may assume t < b. For e > sufficiently small,
at (to + e) lies in a normal neighborhood of a (to), and the segment of a from
348 RIEMANNIAN GEOMETRY [Chap. VII
FIG. 7.22
to to to + e is not a geodesic, since a" {to) 9^ (Fig. 7.22). But then by
Theorem 5.6 its length, L to ,t +t, is strictly greater than the intrinsic dis
tance from a (to) to a (to + e). Thus by the triangle inequality,
L(a) = L a ,t + Lt ,t 0+( + Lt 0+f ,b
> p(p, a(t )) + pM*o), a(t + e)) + p(a(* + e), q)
^p(p,q). 
This result is not too surprising: A shortest road can never turn. Nor
will it have any corners, for a somewhat more sophisticated argument
shows that a (possibly broken) shortest curve must in fact be an (un
broken) geodesic.
We have now reached the main result of this section.
5.9 Theorem Given any two points p and q in a geodesically complete
geometric surface M , there is a shortest geodesic segment from p to q.
Proof. The scheme is an ingenious one, worked out successively by several
mathematicians. (See Theorem 10.9, p. 62, of Milnor [7].) We begin by
selecting a candidate for shortest curve from p to q. Let
P(v) = x(a, v), ^ v ^ 2ir,
parametrize the polar circle C of radius a in a normal neighborhood of p.
By Exercise 6 it follows that the function v — » p(/3(v), q) is continuous on
the closed interval [0, 2w]; hence the function takes on its minimum value
at say vo. Let 7 be the parameter curve, v = vo. Since M is geodesically
complete, y(u) is defined for all u ^ 0. We will show that 7 hits q — in
fact, that
y(r) — q where r = p(p, q). (1)
(This situation is illustrated in Fig. 7.23.) Since 7 has unit speed, it will
follow that
L( T ) == r = p(p, q),
thus proving the theorem.
Sec. 5] LENGTHMINIMIZING PROPERTIES OF GEODESiCS 349
q = y(r)
FIG. 7.23
To establish (1), we use a variant of the standard induction argument
in which integers are replaced by real numbers. For each number u ^
consider the assertion
a(u): p(7(w),q) = r  u (2)
where, as above, r = p(p, q). This says that y (unit speed) is efficient:
After traveling distance u, the distance to q has been reduced by precisely
u. The proof vrill be finished ij we can prove that Gfc(r) is true, for then
p(7(0,q) =0,
so by Exercise 5, y (r) = q. We make a start on this by showing that &(a)
is true for a as above; that is,
p(7(«),q) = r  a. (3)
According to Theorem 5.6, p(p, 7 (a) ) = a; hence by the triangle inequality
r = p(p, q) ^ a + p(7(a), q).
To get (3) we must reverse this inequality. By definition of intrinsic dis
tance, for any e > 0, there is a curve segment a from p to q such that
L{a) ^ p(p,q) + e.
Now a must hit the polar circle C, say at a(to), and we observe that the
portion of a from p to a(t ) has length Li ^ a, and the remainder of a has
length
Lt ^ p(a(t ), q) ^ p(T(a),q).
(The latter since 7(0) was a nearest point to q on C.) Thus
a + p(7(fl), q) ^ Li + L 2 = L(a) ^ p(p, q) + e.
Since e was arbitrary, we obtain the inequality
a + p(y(«),q) ^ p(p, q)
required to prove (3 ) .
350 RIEMANNIAN GEOMETRY [Chap. VII
P = 7(0)
FIG. 7.24
Now we turn to the inductive step of the proof. Since p cannot be nega
tive, Ct(w) is nonsense for u > r. Thus the set of numbers a for which
(1(a) is true has a least upper bound b, with b ^ r. Since the functions
involved in the assertion &(a) are continuous, it follows from the defini
tion of least upper bound that Ct(b) is true.
Here is the plan of the rest of the proof: Assume b < r and deduce a con
tradiction. Then (since b ^ r) we must have b = r; hence Ct(r) is true, as
required.
Let C be a polar circle of radius a* < r — 6 in a normal neighborhood
of y(b). By reproducing the argument for the circle C, we obtain a point
c* such that
p(c*,q) = p( 7 (6),q) a*. (3')
(See Fig. 7.24.) But a (6) reads p(y (6), q) = r  b, so
p(c*, q) = r  b  a*. (4)
The main step that remains is the proof of
c* = y(b + a*). . (5)
This is not too difficult. By the triangle inequality,
p(p,c*) + p(c*, q) ^ p(p, q) = r.
Using (4 ) we get
p(p, c*) ^ b + a*.
But there is a broken curve from p to c* whose length is precisely b + a*.
In fact, referring to Fig. 7.24, we can travel on y from p to y (b) with arc
length b, then from y(b) to c* on a radial geodesic with arc length a*.
Thus by the remark preceding this theorem, this curve is not actually
broken. Hence it is y all the way, so y(b \ a*) is precisely c*.
Finally, we substitute (5) in (4), obtaining
p(y(b + a*), q) = r  (b + a*).
Sec. 5] LENGTHMINIMIZING PROPERTIES OF GEODESICS 351
This says that &(b + a*) is true, and since b + a* is strictly larger than
the upper bound b, we have the required contradiction. 
EXERCISES
1. In the hyperbolic plane H, find the intrinsic distance from the origin
to an arbitrary point p. Deduce that all geodesies of H have infinite
length, hence H is complete. (Hint: Use the triangle inequality.)
2. For the Poincare half plane (Exercise 6 of Section 4):
(a) Find an equation F(x, y, c) = for the routes of the semicircular
geodesies through the point (0, 1).
(b) Find an equation G(x, y, a) = for the polar circles centered at
(0, 1). (Hint: they are the orthogonal trajectories of the curves in
(a).)
(c) Make a sketch showing several curves from each family.
3. At the point p = (r, 0, 0) of the cylinder M: x 2 + y 2 = r 2 , let
d = (0, 1,0) and e 2 = (0,0, 1).
Find an explicit formula for the mapping y (p. 341 ) in this case. What
is the largest normal neighborhood of the point p?
4. Test the scheme used to prove Theorem 5.9 in the special case M = E 2 .
Explicitly: Starting from the geodesic polar mapping
x(u, v) = (pi + u cos v, p 2 + u sin v) at p.
follow the first paragraph of the proof of Theorem 5.9 to determine the
geodesic y.
5. Intrinsic distance is a metric on M. Show that
(a) A normal neighborhood 9l« of p consists of all points q of M such
that p(p, q) < e.
(b) p satisfies the three metric properties: (i) p ^ 0; and p(p, q) =
if and only if p = q, (ii) p(p, q) = p ( q , p ), and (iii) the triangle
inequality.
(Hint: It is necessary to use the Hausdorff axiom for the same
purpose as in the footnote on page 345.)
6. Intrinsic distance is continuous. In a geometric surface M, define p, — > p
to mean that the sequence of real numbers p(p, p.) converges to 0.
Prove that if p t  — > p and q t — ► q, then p(p t , q t ) converges to p(p, q).
7. Let a and /S be two different unitspeed geodesies which start at the
same point a(0) = 0(0). If a and meet again after having traveled
352 RIEMANNIAN GEOMETRY [Chap. VII
the same distance r > 0, that is, o(r) = P(r), prove that neither a or
minimizes arc length past r. (Use the fact that broken geodesies
cannot minimize arc length.)
8. (Continuation). On the cylinder M: x + y 2 = r 2 , prove:
(a) A geodesic starting at (a, b, c) cannot minimize arc length after it
passes through the antipodal line t — * (—a, —b,t).
(b) If q is not on the antipodal line of p, show that there is a unique
shortest geodesic from p to q.
Derive a formula for intrinsic distance on the cylinder.
9. Show that the converse of Theorem 5.9 is false: Give an example of a
geometric surface M such that any two points can be joined by a mini
mizing geodesic, but M is not geodesically complete.
10. Let y: [a, b] — > M parametrize a portion of a meridian of a surface of
revolution M. Prove that y uniquely minimizes arc length. (Hint
Express a competitive curve a as x(ai , a 2 ) where x is a canonical
parametrization, and follow the scheme of Theorem 5.6.)
11. Let M be an augmented surface of revolution (Ex. 12 of IV.l).
(a) If M has only one intercept p (on the axis of revolution) show
that every geodesic y of M starting at p uniquely minimizes arc
length.
(b) If M has a second intercept q show that the assertion in (a) holds
if and only if y does not reach q.
(Hint: No computations are needed.)
6 Curvature and Conjugate Points
We briefly examine the influence of Gaussian curvature K of a geometric
surface M on the geodesies of M.
6.1 Definition A geodesic segment y from p to q locally minimizes arc
length from p to q provided that for any curve segment a from p to q
which is sufficiently near 7 we have L(a) ^ L(y).
To clarify the phrase "sufficiently near," we first define a to be eclose
to y provided there is a reparametrization d of a, on the same interval /
as 7, such that p(a(t), y(t)) < e for all t in I (Fig. 7.25). Then we change
the ending of Definition 6.1 to "provided there exists an e > such that
for any a which is eclose to 7, we have L(a) ^ L(y)." This local mini
mization is strict (or unique) provided we get strictly inequality
L(a) > L(y)
unless a is a reparametrization of 7.
Sec. 6] CURVATURE AND CONJUGATE POINTS 353
FIG. 7.25
To get an intuitive picture of this definition we shall imagine that 7 is
an elastic string — or rubber band — which (1) is constrained to lie in M;
(2) is under tension; and (3) has its end points pinned down at p and q.
Because 7 is a geodesic, it is in equilibrium: If it were not geodesic, its
tension would pull it into a new shorter position. But is the equilibrium
stable? If 7 is pulled aside slightly to a new curve a and released, will it
return to its original position 7? Evidently 7 is (strictly) stable if and only
if 7 is a (strict) local minimum in the sense above, for if a is longer than 7,
its tension will pull it back to 7.
Investigation of local minimization depends on the notion of conjugate
points. If 7 is a unitspeed geodesic starting at p, then 7 is a wparameter
curve, v = v , of a geodesic polar mapping x with pole p. We know that
along 7 the function G = x„« x„ is zero at u = 0, but is nonzero immediately
thereafter (Lemma 5.4). A point 7 (s) = x(s, v ) with s > is a conjugate
point 0/7 (0) = p on 7 provided G(s, v ) = 0. (Such points may or may not
exist.)
The geometric meaning of conjugacy rests on the interpretation of
\/G =  x„  as the rate at which the (radial geodesic) wparameter curves
are spreading apart. Roughly speaking, for fixed e > 0, if y/G =  x„ 
is large, then the distance from x(w, v) to \(u, v + e) is large: The radial
geodesies are spreading rapidly. When \/G is small, this distance is small,
and the radial geodesies are pulling back together again. Thus when G
vanishes at a conjugate point
y(si) = x(si, v ),
it suggests that for v near v , the u parameter curves have all reached this
same point after traveling (at unit speed) the same distance s x (Fig. 7.26).
Unfortunately this meeting may not actually occur. (G controls only the first
derivative terms, and higherorder terms may still be nonzero even though
G vanishes.)
The Euclidean plane E 2 should give the "standard" rate at which radial
354 RIEMANNIAN GEOMETRY [Chap. VII
ty(si) = x(si , » )
P = 7(0)
FIG. 7.26
geodesies spread apart, and for x(u, v) = (u cos v, u sin v), we have
VG = u.
In particular, there are no conjugate points. Let us compare the cases
discussed in Example 5.5, the unit sphere S and hyperbolic plane H. For
2, we find
\/(j = sin u.
Thus radial geodesies starting, say, at the north pole p of 2, spread less
rapidly than in E 2 , since sin u < u for u > 0. Indeed one can see in
Fig. 7.17 that after passing the equator they begin to crowd closer to
gether. All have their first conjugate point after traveling distance x, since
•\/6r(x, v) = sin x = 0. In this case, of course, the meeting of geodesies
actually takes place — at the south pole of 2.
For the hyperbolic plane, we know that the geodesies radiating out from
the origin are just Euclidean straight lines, but they are spreading more
rapidly than in E 2 , as one may surmise from the fact that in H "rulers
shrink as they approach the rim." To prove it we use the data in (2) of
Example 5.5 to compute
\/G = sinh u.
Thus y/G > u for u > 0, and again there are no conjugate points.
6.2 Theorem If 7 is a geodesic segment from p to q such that there are
no conjugate points of p = t(0) on 7, then 7 locally minimizes arc length
(strictly) from p to q.
Proof. Let x be a geodesic polar mapping at p, and restrict its domain to
the region in E 2 on which G > 0. Because there are no conjugate points of
p on 7, we may write y(u) = x(u, v ) for ^ u ^ u . (Thus we allow
u — in this equation, as usual, even though G = there.) Then let a
Sec. 6]
CURVATURE AND CONJUGATE POINTS
355
FIG. 7.27
be another curve segment from p to q with a also defined on the interval
[0, Wo]. Now our proof rests on the fact that if a is sufficiently close to y
(as denned earlier), then a has an expression
a(t) = x(ai(0, aj(0)
which is so close to that of 7 that
ai(0) = 0, / ai(wo) = u .
(Fig. 7.27).
A complete proof of this rather plausible assertion is not trivial. There
is no trouble at u = 0, since we can replace a short initial segment of a
by a radial geodesic — with no loss of generality, since this will not lengthen
a. Then a x and at are constructed in stepby step fashion using the fact
that x is a regular mapping, and hence is locally a diffeomorphism.
Then exactly as in the proof of Theorem 5.6 we have
L(a) = / Vai' 2 + Ga 2 ' 2 dt ^ / 0/ dt
Jo J o
= ai(uo) — ai(0) = uo = L(y)
and if L(a) = L(y), a is merely a reparametrization of 7. 
Our task now is to free the notion of conjugate point from dependence
on geodesic polar mappings. To do so we examine the "spreading coeffi
cient" y/G more closely.
6.3 Theorem Let x be a geodesic polar mapping defined on a region
where G > 0. Then y/G =  x„  satisfies the Jacobi differential equation
(Vg) uu + kVg =
subject to the initial conditions
VG(0, v) = (VG)u(O, v) = 1 for all v.
The restriction G > is needed to ensure that s/G is differentiate.
356 RIEMANNIAN GEOMETRY [Chap. VII
Now \/G(u, v) is actually welldefined for u = 0; indeed,
VG(0,v) = x,(0,») =0.
However, y/G need not be differentiable at u = 0, so we shall interpret
("\/G)u(0, v) and (\ZG) uu (0, v) as limits, for example,
(VG)«(0, v) = lim„^ (y/G) u (u, v).
Proof. The Jacobi equation follows immediately from Lemma 6.3 of
Chapter VI, since as shown in Lemma 5.4, E = 1 and F = f or x. Thus
by the remarks above, it suffices to prove
lim (VG)u(u, v) = 1 (u > 0).
u»0
We need only consider a single radial geodesic y(u) = x(w, v ), setting
g(u) — \/G(u, v ) for u > 0.
Again, since 1? = 1 and F = 0, we obtain a frame field
Ei = 7' = x„, 1?2 = x„/flf on 7 for u > 0.
Since 7 is a geodesic, E\ is parallel and by Exercise 3 of Section 4, so is
Ei. By parallelism, Ei is thus welldefined at u = (Fig. 7.28). Now
Ei(0) = x„(0, v ) = cos v ei + sin v e 2
hence
J5J 2 (0) = —sin vo ei + cos v e 2 .
Furthermore, since E 2 is parallel and x„ = gEi on 7, we get
x„„ = x vu = g E 2 on 7 for u > 0.
^(0)
Sec. 6] CURVATURE AND CONJUGATE POINTS 357
Taking limits as u — > yields
x„„ (0, v ) = (lim g (u) ) E 2 (0 ) .
But
hence
x« (0, v) = cos i> ei + sin v e 2 for all v;
x u »(0, y ) = —sin v ei + cos Voe 2 = Ei(0).
Thus the last equation implies lim u » </(0) = 1; that is,
lim (y/G) u (u, v ) = 1
u»0
for arbitrary v . 
In terms of the spreading apart of radial geodesies, the initial conditions
above show that as they first leave the pole p in any geometric surface,
they are spreading at the same rate as in the Euclidean plane E 2 . For there
we found \/G = u; hence
VG (0, v) = 0, (VG)u(0, v) = 1.
However, the Jacobi equation, written (\/G ! )«« = —K \/G, shows that
thereafter the rate of spreading depends on Gaussian curvature. For K < 0,
radial geodesies spread apart faster than in E 2 . (We observed this earlier
in the hyperbolic plane.) For K > the rate of spreading is less than in
E 2 (as on the sphere).
In particular, to locate conjugate points, it is no longer necessary to
explicitly construct geodesic polar mappings, as we have done heretofore.
We can find y/G on a geodesic 7 by simply solving the Jacobi equation on
7, subject to the given initial conditions. Explicitly, Theorem 6.3 implies
the following result.
6.4 Corollary Let 7 be a unitspeed geodesic starting at the point p
in M . Let g be the unique solution of the Jacobi equation on 7,
g" + K(y)g =
such that gr (0) = 0, </(0) = 1. Then the first conjugate point of 7(0) = p
on 7 (if it exists) is 7 (si), where Si is the smallest positive number such that
g(s t ) = 0.
6.5 Example Conjugate points
(1 ) Let 7 be a unitspeed geodesic starting at any point p of the sphere
S of radius r. The Jacobi equation for 7 is thus g" + g/r 2 = 0, which has
the general solution
358
RIEMANNIAN GEOMETRY
[Chap. VII
g(s) = A sin  + B cos  .
r r
The initial conditions g(0) = 0, ^'(O) = 1, then give g(s) = r sin (s/r).
The first zero of this function with Si > occurs at Si = irr. Thus the first
conjugate point 0/7(0) = p on 7 is at the antipodal point of p. (This agrees
with our earlier computation for the unit sphere by means of geodesic
polar mappings. )
(2) Let 7 be a unitspeed parametrization of the outer equator of a
torus of revolution T with radii R > r > 0. Now 7 is a geodesic and on 7
we know that K has the constant value l/r(R + r). Thus by Corollary
6.4, the first conjugate point y(si) of 7(0) = p on 7 will occur at exactly
the same distance s x along 7 as if 7 were on a sphere with this curvature
K. It follows that si = 7r \/r(R + r).
6.6 Corollary There are no conjugate points on any geodesic in a surface
with curvature K ^ 0. Hence every geodesic segment on such a surface
is locally minimizing.
Proof. Apply Corollary 6.4 to a geodesic 7 in M. Since gr(0) = and
g (0) = 1, we have g(s) ^ for s ^ 0, at least up to the first conjugate
point (if it exists). But K ^ implies g" = — Kg ^ 0, so g is an increasing
function; in fact g ^ 1. Hence g(s) ^ s up to the first conjugate point—
which can thus never occur. The final assertion then follows from Theorem
6.2. I
For example, on a circular cylinder C (K = 0) the helical geodesic 7
from p to q indicated in Fig. 7.29 is indeed stable, as one can verify by direct
experiment. Although locally minimizing, it is certainly not minimizing.
Evidently the straightline segment a provides a considerably shorter
way to get from p to q.
FIG. 7.29
FIG. 7.30
Sec. 6] CURVATURE AND CONJUGATE POINTS 359
To cany the study of conjugate points much further, it is necessary to
use the calculus of variations (see Milnor [7]). We shall quote just one
result, which supplements Theorem 6.2. As soon as a geodesic y starting at
p passes the first conjugate point of p ony, it no longer locally minimizes arc
length. This is fairly easy to see on a sphere 2. In Fig. 7.30 the geodesic y
from p to q is only slightly longer than the first conjugate distance irr.
If the plane of the great circle of y is rotated slightly about an axis through
the end points p and q, it will slice from 2 a curve segment a which, as one
can verify analytically, is strictly shorter than 7. (Note that the only
shorter geodesic from p to q is not near y.)
Theorem 6.3 can also be used to give a rather intuitive description of
Gaussian curvature in an arbitrary geometric surface.
6.7 Lemma If x is a geodesic polar mapping with pole p, then
VG(u,v) =u K(p) \ + o(u) (u ^ 0).
6
Throughout, o(u n ) denotes a function of u and v (u > 0) such that
lim u _o o(u n )/u n = 0. In the formula then, if u is small enough, o(u ) is
negligible compared to the first two terms.
Proof. As before, consider g(u) = y/G (u, v) on a radial geodesic
7 (u) = x(u, v). As a solution of the Jacobi equation on 7, g is differentiate
at u = 0. Thus it has a Taylor expansion
g (u) = 0(0) + g(0)u + g"(0) ~ + </"(0)  + o(u 3 ).
The initial conditions in Corollary 6.4 are g(0) = 0, gr'(0) = 1; hence from
the Jacobi equation we get g" (0) = 0. Differentiating the Jacobi equation
gives
g" + K{y)'g + K(y)g' = 0.
Hence
9" (0) = K(y(0)) = K(p).
Substitution in the Taylor expansion then gives the required result. 
Suppose that the inhabitants of a geometric surface M want to deter
mine the Gaussian curvature of M at a point p. By measuring a short
distance e in all directions fromp, they obtain the polar circle C e of radius e.
Now if M = E 2 , the circumference of C e is just L(C E ) = 2ire. But for
K > the radial geodesies from p are not spreading as rapidly, so C e
should be shorter than 2xe; and for K < they are spreading more rapidly,
so C t should be longer than 2x8.
360
RIEMANNIAN GEOMETRY
[Chap. VII
FIG. 7.31
The relation between L(C e ) and K can be measured with some precision.
For e > small enough, C e is parametrized by v * x(e, v), where x is a
geodesic polar parametrization at p. Thus
L(Ce) = / VG (e, v) dv.
Hence by the preceding lemma,
L(C e ) = 2t (b  K(p) £ + o(e 3 )) .
(*)
Thus if surveyors in M measure L(C e ) very carefully for e small, they can
determine approximately what the Gaussian curvature of M is at p. Taking
limits yields
6.8 Corollary K(p) = lim E ^ (3/7re 3 ) (2x8  L(C e )).
We can easily test formula (*) for a sphere S of radius r in E 3 . As in
Fig. 7.31, the polar circle C e with center p is actually a Euclidean circle
of Euclidean radius r sin &, where & = e/r. Thus by the Taylor series of
the sine function,
UC) = 2, (r si„ E) _ a, (e  ^ + o(e s ))
which gives yet another proof that 2 has Gaussian curvature K = 1/r 2 .
EXERCISES
1. Let x be the polar parametrization of the hyperbolic plane given in
Example 5.5. Derive s/G (m, v) = sinh u in two different ways: by
computing x„ ° x v , and by solving the Jacobi equation.
Sec. 6] CURVATURE AND CONJUGATE POINTS 361
2. If C e is a polar circle around a point p of M, call the region enclosed by
C e the polar disc D e of radius e.
(a) Show that the area of the polar disc is
A(D B ) =7r[e 2 K(p)^ + o(e 4 )]
hence
w , 12 .. ire 2  A(Z)e)
J£(p) = — lim
T e*0 8*
(b) Use this formula to find the Gaussian curvature of a sphere of
radius r.
3. At the origin in the hyperbolic plane, find the length of the polar
circle C e and the area of the polar disc D e (0 < e < 2). Deduce from
each result that K(0) = —1.
4. Let Mbean augmented surface of revolution (Ex. 12 of IV.l).
(a) If M crosses A at only one point p (as on a paraboloid of revolu
tion), show that p has no conjugate points on any geodesic.
(b) If M crosses A at two points, p and q, (as on an ellipsoid of revolu
tion), show that p and q are conjugate on every geodesic joining
them. (Hint: Theorem 6.3 of Chapter V provides a solution of the
Jacobi equation.)
The following exercises deal with a useful variant of the polar geodesic
parametrization in which the pole p is replaced by an arbitrary regular
curve.
5. Let /3: / — » M be a regular curve in M, and let X be a (non vanishing)
vector field on (8 such that /?' and X are linearly independent at each
point. Define
x(w, v) = yx(v)(u);
thus the wparameter curves of x are geodesies cutting across j8 with
initial velocities given by X (Fig. 7.32). Prove:
(a) x is a regular mapping on some region D containing the interval
(0, v), v in /.
(b) By a suitable choice of and X, this parametrization x becomes
(i) the identity map of E 2 (natural coordinates), (ii) the canonical
parametrization of a surface of revolution, and (iii) a ruled para
metrization of a ruled surface (Chapter V, Section 5).
6. (Continuation). If is a umtspeed curve, and X is theunit normal JV
of (Section 4), show that for x: E = 1, F = 0, and \^G is the solution
362
RIEMANNIAN GEOMETRY
[Chap. VII
FIG. 7.32
, Vq), focal point
FIG. 7.33
of the Jacobi equation {y/G) uu + K\/G = such that
VG (0, v) = 1 and (VG)u(O, v) = k„(v).
The natural choice of X in the preceding example means that G for this
parametrization is geometrically significant. If G(it ,v ) = 0, then (by
analogy with conjugate points) we say that x(u ,v ) is a focal point of
along the normal geodesic v = v . Here light rays emerging orthogonally
from 8 tend to meet (Fig 7.33).
7. (a) If ,8 is a circle of latitude on a sphere S, show that the north and
south poles of S are the only focal points of /3.
(b) If fi is a curve in the Euclidean plane, show that its focal points
are exactly its centers of curvature; that is, the points on itsevolute.
(See Ex. 15 of II.4.)
7 Mappings that Preserve Inner Products
We have already seen that a local isometry F: M — > iV carries geodesies
of M to geodesies of N. Using the notation y v for the geodesic with initial
velocity v, we can be more explicit.
7.1 Lemma If F: M — ► N is a local isometry, and v is a tangent vector
to M at p, then
Sec. 7] MAPPINGS THAT PRESERVE INNER PRODUCTS 363
Proof. By the remark above, y = F(y v ) is a geodesic of N. Its initial
velocity is the tangent vector
t'(0) =F*( 7 /(0)) =^*(v)
to N at F(p). Thus by the uniqueness of geodesies (Theorem 4.3), y is
precisely 7f*(«) I
It follows that a local isometry is completely determined by its effect
on just one frame.
7.2 Theorem Let F and G be local isometries from M to N. If for some
one frame ei, e 2 at a point p of M we have
F*M = G*(ei), F*(e 2 ) = #*(«*),
then F = G.
Proof. If M is geodesically complete, the proof is particularly easy. If
q is an arbitrary point of M, then by Theorem 5.9 there is a vector v at
the special point p such that y v (r) = q. From the hypothesis on F* and
G* , we deduce by linearity that F# and G* agree on v = Cid + c 2 e 2 . Thus
the preceding lemma shows that
F(y v ) = 7**<») = 7o*(«) = #(7«)
Hence, in particular,
F(q) = F(y v (r)) = G(y v (r)) = G(q)
or all points q of Af .
The proof for M arbitrary is a refinement. Using Lemma 5.3 it is possible
to get a broken geodesic from p to q and deduce F (/3) = 0(/8) by applying
the argument above to each unbroken segment of 0. 
We shall now use the fact that local isometries preserve geodesies to
construct some local isometries. The goal is to demonstrate a family re
semblance among geometric surfaces of the same constant curvature.
Given any number K, there is a particularly simple geometric surface
M(K) whose Gaussian curvature has constant value K.
(1) If K > 0, let M (K) be the sphere S of curvature K (hence radius
i/Vk).
(2) If K = 0, let M(K) be the Euclidean plane E 2 .
(3) If K < 0, let M(K) be the hyperbolic plane H of curvature K
(hence pseudoradius \/y/ — K: see Exercise 4 of Section 2).
We shall call M(K) the standard geometric surface of constant curvature
K. Of course, there are many other constant curvature surfaces; these are
distinguished by the fact that they are geodesically complete and simply
connected (p. 176).
364 RIEMANNIAN GEOMETRY [Chap. VII
7.3 Theorem Let N be a geodesically complete geometric surface with
constant Gaussian curvature K. Then there is a local isometry F of the
standard surface M(K) onto N.
The first mapping in Example 4.6 of Chapter VI is an instance of this
theorem, as is the local isometry (Exercise 6 of Section 2) of a sphere onto
a projective plane.
Proof. The Case K < 0. We use the language of Example 2.4, where
K = — 1. A mere change of scale (Exercise 4 of Section 2) takes care of arbi
trary K < 0. As in (2) of Example 5.5, let p be the origin of H = M ( 1 ),
with ei = £A(p) and e 2 = £Mp). Let £i, e 2 be a frame at an arbitrary
point of N. Then let x and x be the resulting geodesic polar mappings of
H and N.
For the surface N, we assert that
(1) x is defined on the entire right half plane S: u > (a consequence
of geodesic completeness).
(2) Its image x(S) covers all of N except possibly the pole p (a conse
quence of Theorem 5.9 and the definition of geodesic polar mappings).
(3) x: S — > N is a regular mapping. (By Lemma 5.4, E = 1, and F = 0,
but we have observed earlier that the Jacobi equation for K = — 1 gives
VG = sinh u, so EG  F 2 = sinh 2 w > on S.)
This general result is thus valid for x: S — » H as well, but here we know
more. By Example 5.5 the whole surface H is a normal neighborhood of
the pole p; thus x has only the usual ambiguities of polar coordinates; the
equation x(u, v) = q determines u uniquely, and v uniquely but for addi
tion of some multiple of 27r (q ^ p). From this extra information we con
clude that the formula
F(x(u, v)) = x(u, v)
is consistent and thus defines a mapping F of H onto all of N.
(To prove differentiability of F at the pole p we must resort, as in the
proof of Lemma 5.3, to the mappings y and y corresponding to x and x.)
It is easy to show that F is a local isometry by using the criterion of Lemma
4.5 of Chapter VI. Indeed, from (3) above, we have
E = 1 = E, F = = F, G = sinh 2 w = G, for u > 0,
and at the pole p the preservation of inner products is an honest consequence
of continuity.
The Case K = 0. This is a wordforword copy of the preceding argument
except that
M(K) = E 2 and G = G = u.
Sec. 7] MAPPINGS THAT PRESERVE INNER PRODUCTS 365
The Case K > 0. Here a new idea is required, since the largest normal
neighborhood 91 of a point p in the sphere 2 = M(K) is not all of 2: The
antipodal point — p is omitted.
Arguing as in the case K < 0, we get a local isometry Fi: 91 — > N. Now
repeat this argument once more at a point p* in 2 different from both p
and —p. We obtain another local isometry Fz: 91* — > N where 91* is all
of 2 except — p*. The frames determining F 2 are chosen so that the deriva
tive maps of Fi and F 2 agree at p*. Thus by Theorem 7.2, F\ and F 2 are
identical on the overlap of 91 and 91*. But 91 and 91* cover the whole sphere
2, so taken together F x and F 2 constitute a single local isometry F: 2 — ► N.
Because 2 is compact and N is connected, Exercise 6 of Section 7 in Chapter
IV shows that F carries 2 onto M. 
An isometry F: M —> M of a geometric surface onto itself may be viewed
as a symmetry of M . Every feature of the geometry of M is the same at each
point p as at F(p), since this geometry consists of isometric invariants.
The results of Exercise 9 of Section 4, Chapter VI, show at once that the
set ${M ) of all isometries F: M —> M forms a group, just as do the set of
all isometries of Euclidean space (Exercise 7 of Section 1 of Chapter III).
We call d(M) the isometry group of M.
This group d{M) is, of course, intrinsic to M, and when M is a surface
in E 3 , should not be confused with the group S(M) of Euclidean symmetries
of M (Exercise 7 of Section 8, Chapter VI). A Euclidean symmetry F of
M C E 3 is an isometry of E 3 such that ¥{M) = M; these exist when the
shape of M in E is symmetric in the ordinary sense of the word. Each
Euclidean symmetry F of M gives rise to an isometry F  M: M — > M, but
in general this process does not give all isometries of M C E 3 (Exercise 9).
For an arbitrary geometric surface M, the isometry group &(M) gives
a novel algebraic description of M. Roughly speaking, the more sym
metrical M is the larger d(M ) is. For example, although we shall not carry
out the proof, the ellipsoid
M: W> + t + t = l (a>b> c)
a z b z c l
has exactly eight elements in its isometry group, these all arising from its
Euclidean symmetries as described above: three reflections (one in each
coordinate plane), three 180° rotations (one around each coordinate axis),
the isometry p — ► — p, and, of course, the identity map of M .
The smallest possible isometry group d(M) occurs when the identity
map of M is the only isometry of M. We can produce such a geometric
surface by putting a bump on the ellipsoid in such a way as to destroy all
seven of its nontrivial isometries.
By contrast, a geometric surface M has the maximum possible sym
366
RIEMANNIAN GEOMETRY
[Chap. VII
metry when every possible isometry permitted by Theorem 7.2 actually
exists. That is, given frames ei, e 2 and ei , e 2 at any two points of M , there
exists an isometry F: M —*■ M such that
F*(e x ) = ei, F*(e 2 ) = e 2 .
In this case, we shall say that M is framehomogeneous; any two frames on
M are symmetrically positioned.
Thus what we proved in Theorem 2.3 of Chapter III is that E 3 is frame
homogeneous, and the same proof is valid for arbitrary E ra , in particular
for E . In the exercises for this section, we shall see that every standard
surface M(K) of constant curvature is framehomogeneous.
7.4 Definition A geometric surface M is pointhomogeneous (or merely
homogeneous) provided that given any two points p and q of M there is an
isometry F: M — > M such that F(p) == q.
A framehomogeneous surface, is of course, homogeneous — but not
conversely. A circular cylinder C in E furnishes an example. In fact, if
F is a rotation of E 3 about the axis of C, or a translation of E 3 along this
axis, then F carries C onto C, producing an isometry of C. Hence given any
points p and q of C, we can first rotate to bring p to p on the same ruling
as q, then translate p to q. The composition of these two isometries is an
isometry carrying p to q. On the other hand, C is not framehomogeneous:
all its points are geometrically equivalent, but not all its frames. (Proof:
For the unit vectors shown in Fig. 7.34, no isometry could carry ei to ei,
for by Lemma 7.1, F would have to send the onetoone geodesic y e to
the periodic geodesic 7^ — an impossibility, since F is onetoone.)
Homogeneity is a very strong restriction.
7.5 Theorem If a geometric surface M is homogeneous, then M is
geodesically complete and has constant Gaussian curvature.
Proof. Constancy of curvature follows immediately from the definition
of homogeneity and the fact that isometries preserve curvature. The proof
Sec. 7] MAPPINGS THAT PRESERVE INNER PRODUCTS 367
F*(u) = a'iU)
FIG. 7.35
of completeness is more interesting. If M is not geodesically complete, there
is a maximal unitspeed geodesic a defined only on an interval, say, /:
t < a, which is not the whole real line. Let us show that this is impossible.
By Lemma 5.3, all geodesies emanating from some arbitrary point p of
M run at least for some fixed distance e > 0. Choose t in I such that
a — to < e/2. Because M is homogeneous, there is an isometry F: M — *• M
such thatF(p) = a(t ). Now for some unit vector u at p, F* (u) = a {to).
Thus the geodesic segment F(y u ) has initial velocity
F*(yJ(0)) = F*(u) = a (k)
and runs for distance e at unit speed (Fig. 7.35). But then a shift of para
metrization enables us to apply Theorem 4.3 and thereby define a on the
interval I*: t < t + e. But t + e > a, so this contradicts the maximality
of the interval /, and thus proves M is geodesically complete. 
As the title of this section suggests, (local) isometries are not the only
innerproductpreserving mappings of importance in geometry. We shall
take a brief look at the other main types.
7.6 Definition Let F: M — > E 3 be a mapping of a geometric surface into
E . If the derivative map F* preserves inner products of tangent vectors,
then F is an isometric immersion. If F is also onetoone, then F is an
isometric imbedding. An isometric imbedding F such that the inverse func
tion F~~ : F(M) — > M is continuous is said to be proper.
This definition is unduly restrictive. Evidently we could replace E 3 — or
even M — by any Riemannian manifold (p. 308).
7.7 Lemma If F: M — * E 3 is a proper isometric imbedding of a geometric
surface M into E 3 , then the image F(M) is a surface in E 3 and the function
F: M — * F(M) is an isometry.
Proof. If x: D — ► M is a proper patch in M, then the composite mapping
F(x) : D — » E 3 is a patch that lies in F(M ). Furthermore, F(x) is a proper
patch. In fact, its inverse function F(x(D)) —> D is just x 1 /^ 1 , which is
continuous since x 2 and F' 1 are continuous. Thus we can easily check
368 RIEMANNIAN GEOMETRY [Chap. VII
Definition 1.2 of Chapter IV. Now as a geometric surface, F(M) uses the
dot product of E , and by definition F: M — > E 3 preserves inner products.
Hence when considered as a mapping of M onto F{M), F preserves inner
products. 
Thus the study of the geometry of surfaces in E 3 is exactly the same as
the study of proper isometric imbedding s of geometric surfaces into E 3 .
This rather technical fact is important only because it suggests a considera
ble generalization of our work in Chapters V and VI. We could just as well
have studied the far larger class of isometric immersions into E 3 , dropping
the onetoone and properness restrictions. There is, in fact, no real diffi
culty involved, except for complications of notation.
As in the special case discussed on page 308, the image F(M) of an
isometric immersion F: M — > E 3 may cut across itself; nevertheless, we
shall think of it as a kind of defective surface in E 3 . If we define the shape
operator of such an immersed surface, this should at least suggest how to
generalize the rest of Chapters V and VI.
Because inner products are preserved, an isometric immersion F is
regular. Thus F*(T P (M)) is a twodimensional subspace of 7V( P )(E 3 ); it
plays the role of a tangent plane for F(M) at F(p). A unit normal function
U assigns to each point p (in some region of M ) a unit vector orthogonal
to F* (T P (M)). If a is a curve in M, then U a is a vector field on F(a) in
E 3 . Then if v is the initial velocity of a, we define S(\) to be the unique
vector in T P (M) such that
F*(S(v)) =  Uj(0)
This shape operator S is again a symmetric linear operator on T P (M).
Most of our earlier results hold up rather well under generalization.
For example, if the Gaussian curvature K of M is defined intrinsically as
in Section 2 — then by reorganizing the logic in Chapter VI, Section 2, we
can show that K — det S.
A theorem such as Theorem 3.7 of Chapter VI becomes more informative:
If M is a compact surface with constant curvature K(>0), and F: M —*■ E 3
is an isometric immersion, then F is an isometry of M onto a Euclidean
sphere 2 of radius 1/s/K in E .
In other words, even if we give F(M) permission to cut across itself,
this cannot happen: F{M) can only be an ordinary round sphere in E .
We have seen that there are geometric surfaces M which cannot be
isometrically imbedded in E 3 — for example, the flat torus (Example 2.5)
or the projective plane, Exercise 6 of Section 2. In this case it is natural
to try to imbed M in a higher dimensional Euclidean space E n . The larger
n is, the less difficult the task becomes. (Roughly speaking, with more
dimensions available for M to curve through, there is a better chance that
Sec. 7] MAPPINGS THAT PRESERVE INNER PRODUCTS 369
a shape can be found for M which is compatible with its intrinsic geometry.
See Chapter VI, Section 9.)
Thus, although there are no flat toruses in E 3 , they can be found in E 4 .
7.8 Example An isometric imbedding of a flat torus in E 4 . Start with
the mapping x: E 2 — > E 4 such that
x(u, v) — (cos u, sin u, cos v, sin v).
If x is the parametrization of the flat torus T given in Example 2.5, then
the formula
F(x(u, v)) = x(u, v)
is consistent; in fact it defines a onetoone mapping F: T — > E 4 . The proof
consists in observing that
x(u, v) = x(wi, vi) <=>t*i = u + 27rra,i>i = v + 2ira <=> \(u, v) = x(u u t> x ).
Reading the implication arrows from left to right we get the required
consistency; the reverse direction shows that F is onetoone.
Then F is an isometric imbedding provided F* preserves inner products.
In the usual fashion we compute
x u = ( — sin u, cos v, 0, 0)
x„ = (0, 0, —sin v, cos v)
Hence
E = l, F = 0, (5=1.
These functions agree with E, F, and G for x, so exactly the same argument
used to prove Lemma 4.5 of Chapter VI shows that F * preserves inner
products.
The general situation here is not well understood. Although every
compact geometric surface can be isometrically imbedded in E , it re
mains a possibility that 17 can be replaced by as low a dimension as 4.
EXERCISES
1. Let F: M — ► N be a local isometry, and suppose that M is geodesically
complete. Show that F is onto if and only if N is geodesically complete.!
2. Prove that a geodesically complete geometric surface with constant pos
t Though the proof is not elementary, it is known that both of these properties
are consequences of the geodesic completeness of M.
370 RIEMANNIAN GEOMETR\ [Chap. VII
itive curvature is compact. (The result still holds if merely K ^ c > 0.
See Myers' theorem in Hicks [5].)
3. Suppose that in M any two points can be joined by at least one geodesic,
and that in N any two points can be joined by at most one geodesic.
Prove that every local isometry F: M — * N of such surfaces is oneto
one.
4. Let F: M — » M be an isometry that is not the identity mapping. If a
unit speed curve is fixed under F, that is,
F(a(s)) = a(s) for alls,
show that a is a geodesic of M.
5. Let x and x be geodesic polar parametrizations of normal neighbor
hoods 9l E and 9l E (same e) in two geometric surfaces. If K(x) = K(x)
on the common domain S e of x and x, prove that 9l e and 3l E are iso
metric.
6. Prove that the sphere S and hyperbolic plane H are frame homogene
ous. (Hint: for 2 derive the required isometries from orthogonal
transformations of E 3 ; fori/ use Theorem 7.3 and a preceding exercise.)
7. Show that the flat torus (Example 2.5) is homogeneous, but not frame
homogeneous, and that an ordinary torus of revolution in E 3 is not
homogeneous.
8. Prove:
(a) For the right circular cylinder C: x 2 + y 2 = r 2 in E 3 , every isometry
F: C > C has the form
^(p) = (Pi cos & ± p 2 sin #, p x sin # ± p 2 cos t?,ep 3 + a)
where e = ±1.
(b) Every isometry of a sphere or right circular cylinder in E 3 is the
restriction of an isometry of E 3 .
9. Let M be the cylinder in E 3 whose crosssectional curve is the ellipse
4a; + y 2 = 4. (Any other closed noncircular curve could be used.)
Show that there is an isometry of M which is not the restriction of an
isometry of E 3 . (Hint: parametrize M by x(u, v) = a(u) + vU 3 ,
where a is a periodic unitspeed parametrization of the ellipse.)
10. In the sphere 2 of radius r, let T be a triangle whose sides are geodesic
segments of lengths a, b, and c (all less than vr). Let # be the angle
of T at the vertex p opposite side a.
(a) Prove the law of cosines:
a b c , . b . c
cos  = cos  cos  + sin  sin  cos #.
Sec. 7] MAPPINGS THAT PRESERVE INNER PRODUCTS 371
(b) Show that this formula approximates the usual Euclidean law
of cosines when r is large compared to a, b, c.
{Hint: to determine cos # find unit vectors 115, u c at p tangent
to the sides b and c. )
11. Prove that the projective plane (Exercise 6 of Section 2) is frame
homogeneous. (Hint: if F : 2 — > 2 is an isometry of the sphere 2 C E 3 ,
then F( — p) = — ^(p), hence there is a mapping F: 2 — > 2 such that
PF = FP.)
12. Show that the isometry groups of isometric surfaces are isomorphic.
13. If M is a surface in E 3 that does not lie in a plane, show that the func
tion F —> F  M is an isomorphism of the Euclidean symmetry group
S(M) onto a subgroup of the isometry group d(M).
Isometries of the hyperbolic plane may be constructed explicitly by
recognizing a point of the plane as a complex number
z = u + iv = (u, v),
and using exercises of Section 1. Thus if  z  denotes the magnitude of z,
I I 2  2,2
\z\ = zz = u + v ,
the hyperbolic plane may be described as the disc  z \ < 2 with conformal
geometric structure given as in Example 1.3 by g(z) = 1 — \z\ 2 /4.
1 4. (A translation of the hyperbolic plane. ) For a fixed real number c = (c, )
in H, let T be the mapping T(z) = 4 [(z + c)/(cz + 4)] defined on H.
(a) Show that T(H) <z H, and that T: H — > H is onetoone and onto.
If H denotes the same disc, \z\ < 2, but with the usual Euclidean
structure, then Exercise 7 of Section 1 shows that T: H' — ► H' is a
conformal mapping with scale factor X (z) =  dT/dz  .
(b) Show that this scale factor is
X(z) = 4
4  c z
cz + 4  2
(c) Deduce that T: H — > H is an isometry of the hyperbolic plane.
(Hint: Use Ex. 9 of VII. 1.)
These methods can be used to show that H is framehomogeneous,
and — carried somewhat further — to give an elegant derivation of the
geodesies of H.
15. (The Poincare half plane P is isometric to the hyperbolic plane H.)
In terms of complex numbers, P is the halfplane Im z > with con
formal geometric structure g(z) = Im z. (Im z is the imaginary part
v of z = u + iv. ) Let F be the mapping
372 RIEMANNIAN GEOMETRY [Chap. VII
F(z) = Z + 2i
{Z) iz + 2>
defined on H. Show that
(a) Im F(z) = (4 \ z\ 2 )/ \iz + 2 \ 2 .
(b) F is a onetoone mapping of H onto P. (Compute F _1 explicitly.)
(c ) Relative to Euclidean structures, F is conformal, with scale factor
X (z) = 4/  iz + 2  2 .
(d) F: i/ — ► P is an isometry.
Make a sketch of i7 and P indicating the images in P of each of the
four quadrants of H.
8 The GaussBonnet Theorem
We have seen that the Gaussian curvature K of a geometric surface M
has a strong influence on other geometric features of M such as parallel
translation, geodesies, isometries, and, of course, the shape of M if it
happens to be in E . Now we will show that the influence of Gaussian curva
ture penetrates to the ultimate topological conformation of M — to proper
ties completely independent of the particular geometric structure on M.
The main step in this proof is a theorem which relates the total curva
ture of a 2segment to the total amount that its boundary curve turns.
For an arbitrary curve a in M, the geodesic curvature tells its rate of
turning relative to arc length. Thus to find the total amount a turns, we
integrate with respect to arc length :
8.1 Definition Let a: [a, b] — > M be a regular curve segment in an oriented
geometric surface M. The total geodesic curvature /« k ds of a is
.*(b)
K g (s) ds
•'s(a)
where k (s) is the geodesic curvature of a unitspeed reparametrization of
a.
The total geodesic curvature of a in M is thus an analogue of the total
Gaussian curvature of a surface in E 3 . For example, let C be a circle of
radius r in E 2 , where E 2 has its natural orientation. If a is a curve making
one counterclockwise trip around C, then a has constant geodesic curvature
Kg = l/r. Thus
/ Kg ds =  2nr = 2t,
K r
regardless of the size of the circle. A clockwise trip around C will have total
Sec. 8] THE GAUSSBONNET THEOREM 373
curvature — 2ir, for in general: If the orientation of M is kept fixed, then
the total geodesic curvature of a curve segment a is unaffected by orien
t&tioiipreserving reparametrization, but has its sign changed by an orien
t&tionreversing reparametrization. (The former is clear from the defini
tion; the latter can be deduced, for example, from the following lemma.)
8.2 Lemma Let a: [a, b] — *■ M be a regular curve segment in a region of
M oriented by a frame field E u E%. Then
J Kg ds = <p(b) — <p(a) + J
«12
where <p is an angle function from Ei to a on a, and w^ is the connection
form of E x , E%
Proof. None of these terms are affected by orientationpreserving
reparametrization; thus we may assume that a is a unitspeed curve. But
then the result follows immediately by integrating the formula in Lemma
4.5 
For the integration theory in Chapter VI, Section 7, we used 2segments
x: R — > M which were onetoone and regular on the interior R° of R.
Now we shall impose the more stringent requirement that x be onetoone
and regular on the boundary of R as well. (This is equivalent to saying
that x: R — *• M is the restriction to R of a patch defined on some open set
containing R.)
When x is a onetoone regular 2segment, its edge curves a, 0, y, 8
(Definition 6.4, Chapter IV) are onetoone regular, and we shall think
of the boundary dx = a + (i — 7 — 5 as a single broken curve enclosing
the rectangular region x(R). We now want to define the total geodesic
curvature of dx. The definition of geodesic curvature shows that the total
geodesic curvature of a curve is simply the total angle that its unit tangent
T turns (relative to arc length). But to go all the way around
dx = a \ & — 7 — 5
we must get not merely the total turning on the edge curves; that is,
/ Kg dS = I Kg dS { I Kg dS + I Kg rfS + /
= / Kg dS + / Kg dS — f Kg dS — IK
J a JB Jv J S
Kb
&
ds
s
but also the angles through which a unit tangent T on dx would have to turn
at the four corners of the rectangular region x(R) (Fig. 7.37). For
R: a ^ u ^ b, c ^ v ^ d
374
RIEMANNIAN GEOMETRY
[Chap. VII
0(0)
«'(*)
FIG. 7.36
FIG. 7.37
these "corners"
pi = x(a,c), pa = x(b,c), p 3 = x(6,d), p 4 = x(a,d)
are called the vertices of x(R).
In general, if a regular curve segment a in an oriented region ends at the
starting point of another segment /3, say a (I) = /8(0), then the turning
angle e from a to /3 is the oriented angle from a (1 ) to /3' (0) which is smallest
in absolute value (Fig. 7.36). For a 2segment we use the orientation
determined by x, that is, the area form dM such that dM(x u ,x v ) > 0, to
establish some terminology which is familiar in the case of a polygon in
the plane.
8.3 Definition Let x: R — » M be a onetoone regular 2segment, with
vertices pi, p2, p3, p 4 . The exterior angle ey of x at py (1 ^ j " ^ 4) is the
turning angle at py derived from the edge curves a, 0, —7, —8, a, • • • in
order of occurrence in dx. The interior angle ty of x at py is t — ey (Fig. 7.37).
This definition is given with more general applications in mind; in the
case at hand, exterior angles can easily be expressed in terms of the usual
coordinate angle 1? from x„ to x„ (0 < & < it) by
81 = X — #1 82 = #2 83 = 7T — #3 84 = t?4
where #y is the coordinate angle at the vertex py. For example, let us con
Sec. 8] THE GAUSSBONNET THEOREM 375
FIG. 7.38
sider the situation at p 3 , as shown in Fig. 7.38. By the definition of the
edge curves, ($' is x„, but (—7)' is — x u , since —7 is an orientationreversing
reparametrization of 7. Thus e 3 + #3 = ir. (Analytical proofs may be
based on the definition of oriented angle in Section 7, Chapter VI.)
We are now ready to prove the fundamental result of this section.
8.4 Theorem Let x: R — > M be a onetoone regular 2segment in a
geometric surface M. If dM is the area form on x{R) determined by x,
then
ff KdM + f Kg ds + (ex + e 2 + e 3 + e 4 ) = 2*
*• v ^ v N/ , . ^
total Gaussian total geodesic
curvature of x curvature of 3x
(The geodesic curvature and exterior angles use the orientation of x(^)
given by dM, where dM (x„,x„) > 0. Note that M itself need not be ori
ented — or even orientable.)
We call this result the GaussBonnet formula with exterior angles. Since
Bj = ir — ij for 1 ^ j '^ 4, the formula may be rewritten
f[ KdM + f Kg ds = (n + i 2 + 13 + i 4 )  2tt
in terms of the interior angles oix(R).
Proof. Let Ei = x u /%/E on the region x(R). Then let E 2 be the unique
vector field such that E h E 2 is a frame field with dM '(Ei, E 2 ) = +1. In
this case (compare with p. 292), the second structural equation becomes
dun = Kdi a 2 = K dM
The power for this proof is supplied by Stokes' theorem (6.5 of Chapter
376 RIEMANNIAN GEOMETRY [ Chap . V 
IV) which gives
// KdM + f co 12 = 0. (1)
"^ " x « J Let us now use Lemma 8.2 to evaluate
/ OJ12 = / CO12 + / 0)12 — / C012 — / C0l 2 . (2)
FIG. 7.39 3x J « J * J y _ Js
On a we have a = x u = y/E E u so the
angle <p from E\ to a is identically zero. Thus by Lemma 8.2,
/ wi2 = / K g ds. (3)
J a J a
Next we try a harder case, say J s o>i 2 . Here the angle <p from
El = VE toS ' = Xv
is precisely the coordinate angle # from x M to x v (See Fig. 7.39.) Hence by
Lemma 8.2 we get
/ k ds = & 4 — #i + f
OJ12
where, as above, < & 3 < t is the coordinate angle at the vertex p y of x
(1 ^ j ^ 4). But since
#i = ir — ei and #4 = 8 4 ,
This becomes
/ W12 = t — 81 — 84 + / k<7 ds. (4)
In an entirely similar way we find
/ 0>12 = — IT + 8 2 + 83 + I Kg dS (5)
and
J cow = J K fl ds. (6)
Thus (2) becomes
/ «u = / Kg ds \ I Kg ds — / Kg ds — I Kg ds — 2tt + (ei + e 2 + e 3 + e 4 )
•'ax •'a «/0 J T Jj
= / k ds — 2x + (ei + e 2 + e 3 + 84).
Substitution in (1) then yields the required formula. 
Sec. 8]
THE GAUSSBONNET THEOREM
377
FIG. 7.40
The GaussBonnet formula actually depends not on the particular
mapping x: R — » M, but only on its image (R = x(R). Explicitly, if x is
another onetoone regular 2segment x with the same image (R, then each
of the six terms in the GaussBonnet formula for x has exactly the same
numerical value as the corresponding term for x. This should be no sur
prise if x and x have the same orientation, that is, determine the same
area form on (R. But suppose they have opposite orientation (as in Fig.
7.40) so that dM x = —dM x . To take the trickiest case, consider correspond
ing edge curves such as a. and in the Fig. 7.40. Now a and /8 run in
opposite directions: /8 is an orientationreversing reparametrization of a.
But the geodesic curvatures of a and j8 are computed in terms of the op
posite area forms dM x and dM x . Thus there are two sign changes, so
/. Ka ^ = i
ds.
The remainder of this section will be devoted to applications of the Gauss
Bonnet formula. The basic idea is to extend it to more general regions —
in particular to entire geometric surfaces. To do so we must look at some
fundamental properties of surfaces which do not involve geometry.
A rectangular decomposition 3D of a surface M is a finite collection of
onetoone regular 2segments xi, • • • , x/ whose images cover M in such
a way that if any two overlap they do so in either a single common vertex
or a single common edge.
Evidently a rectangular decomposition is a special kind of paving (Defi
nition 7.3 of Chapter VI), but the regions Xi(Ri) are now really "rectangu
lar" (since x» is onetoone regular on all of Ri) and they are required to
fit together very neatly, as in Fig. 7.41 (compare the paving in Fig. 6.17).
8.5 Theorem Every compact surface M has a rectangular decomposition.
(Hence in particular M has a paving.) This result is certainly plausible,
for if M is made of paper, we could just take a pair of scissors and cut out
rectangular pieces until all of M is gone. A general proof is given in Lef
schetz [8] (use Exercise 10).
We shall understand that a rectangular decomposition 3D carries with it
not only its rectangular regions Xi(Ri) — called faces — but also the vertices
and edges of these regions.
378
RIEMANNIAN GEOMETRY
[Chap. VII
FIG. 7.41
8.6 Theorem If 3D is a rectangular decomposition of a compact surface
M , let v, e, and / be the number of vertices, edges, and faces in D. Then
the integer v — e + / is the same for all rectangular decompositions of M .
This integer x(M) is called the EulerPoincare characteristic of M.
The natural proof of this famous theorem is purely topological, however
it is an easy consequence of Theorem 8.8.
These results may easily be generalized. First, instead of an entire surface,
we could deal with a polygonal region, one which can be decomposed into
rectangular regions Xi(Ri) fitting together neatly (as above). Second,
we could everywhere replace rectangles R by polygons. (A polygon P is the
bounded region in E enclosed by a simple polygonal curve — P including
this curve.) Combining both generalizations we get the concept of polyg
onal decomposition 3D of a (polygonal) region (R in M. The EulerPoincare
characteristic x(#0 of (R is still independent of the choice of polygonal
decomposition 3D.
8.7 Example EulerPoincare characteristic.
(1) A sphere 2 has x(2) = 2. By "inflating" a cube as in Fig. 7.42 we
get a rectangular decomposition 3Di of 2. SDi has v = 8, e = 12, / = 6 —
and thus x = 2. Inflating a prism gives a polygonal decomposition SD 2
with v = 6, e = 9, / = 5— again x = 2 (Fig. 7.42).
^l^
Sec. 8] THE GAUSSBONNET THEOREM 379
M + H = M'
FIG. 7.43
(2) A torus T has x(T) = 0. Picture T as a torus of revolution, and cut
it along any three meridians and three parallels. This gives a rectangular
decomposition 2D for which v = 9, e = 18, / = 9, hence x = 0.
(3) Adding a handle to a compact surface reduces its EulerPoincare*
characteristic by 2.
Roughly speaking, a "handle" is a torus with the interior of one face
removed. (We suppose that M and the torus are given in some rectangular
decomposition.) To add a handle to M, remove the interior of a face of M
also, and to the resulting rim smoothly attach the rim of the handle, so
that the vertices and edges of the two rims coincide (Fig. 7.43).
This operation produces a new surface M already equipped with a rec
tangular decomposition. The EulerPoincare" characteristic of M' is
x(M)  2,
since its decomposition has exactly two faces less than M and the torus
combined. (Coalescing the two rims eliminates four vertices and four
edges as well, but this has no effect on x)
It is easy to see that diffeomorphic surfaces have the same EulerPoincare
characteristic, for if x x , • • • , x/ is a decomposition of M and F: M — > M
is a diffeomorphism, then F(xi), • • • , F(x f ) is a decomposition of M with
exactly the same v, e, and /.
For example, no matter how wildly we distort the sphere
2:z 2 + ?/ 2 + z 2 = 1,
the resulting surface M will still have EulerPoincare" characteristic 2.
So long as no geometric structures are involved, the word "sphere" might
be used to mean "a surface diffeomorphic to 2." To avoid any possible
confusion we shall stay with the longer terminology.
Suppose we start with the sphere 2 and successively add h handles
(h = 0, 1, 2, • • • ) to obtain a new surface 2 (ft). What is remarkable about
the operation of adding handles is that every compact orientable surface M
is diffeomorphic to some 2 (ft). In this case we shall say that M itself has
ft handles. By (3) of Example 8.7, it follows that
x(M) = X (2(ft)) = x (2)  2ft = 2  2ft.
380
RIEMANNIAN GEOMETRY
[Chap. VII
FIG. 7.44
In Fig. 7.44, for example, all four surfaces have just one handle, and thus
all have x = 0.
Although we have used the concepts of calculus in this brief discussion
of the EulerPoincare" characteristic, our remarks remain valid if differen
tiability is everywhere replaced by continuity. The EulerPoincare* char
acteristic is, in fact, a topological invariant.^
Returning now to the subject of geometric surfaces, we prove a spectacu
lar consequence of Theorem 8.4.
8.8 Theorem (GaussBonnet) If M is a compact orientable geometric
surface, then the total Gaussian curvature of M is 2irx(M), where xiM)
is the EulerPoincare* characteristic of M.
Proof. Fix an orientation of M with area form dM, and let 3D be a rec
tangular decomposition of M whose 2segments x u • • • , x f are all positively
oriented. Thus 2D is in particular an oriented paving of M as denned in
Chapter VI, Section 7. By definition the total curvature of M is
// KdM = ±J[
KdM
(1)
We shall apply the GaussBonnet formula to each summand. (This is
permissible, since on each region Xi(Ri) the area form dM is the one de
termined by x;. ) In terms of interior angles, this formula reads
t A topological invariant is a property preserved by every homeomorphism (that
is, continuous function with a continuous inverse). A diffeomorphism is a homeo
morphism, but not conversely. However it is a peculiarity of low dimensions that
two surfaces are diffeomorphic if (and only if) they are homeomorphic.
Sec. 8]
THE GAUSSBONNET THEOREM
381
// K dM =  / Kg ds 2x + ( 4l + i 2 + t 3 + l 4 )
(2)
Now consider what happens when (2) is substituted in (1).
Because M is a surface— locally like E 2 — each edge of the decomposition
£> will occur in exactly two faces, say Xi(Ri) and xj(Rj). Let en and a,
be the parametrizations of this edge occurring in dx» and dx h respectively.
Because these regions have the same orientation as M itself, ai and « y
are orientationreversing reparametrizations of each other, as in Fig.
7.45. Thus
/ Kg dS + / Kg dS = 0.
It follows that
»=1 •'dx.
Kg ds =
(3)
for we have just seen that the integrals over edge curves will cancel in
pairs. (As usual, we write v, e, and / for the number of vertices, edges,
and faces in the decomposition.)
Thus the substitution of (2) in (1) yields
I
KdM = 2x/+ tf
(4)
where d is the sum of all interior angles of all the faces in the decompo
sition. But the sum of the interior angles at each vertex is just 2x (Fig.
7.46), so d = 2irv. Thus
I
KdM = 2tt/ + 2ttv
(5)
FIG. 7.45
382 RIEMANNIAN GEOMETRY [Chap. VII
FIG. 7.46
A simple combinatorial observation will complete the proof. The faces
of the decomposition 3D are rectangular: Each face has four edges. But each
edge belongs to exactly two faces. Thus 4/ counts e twice; that is, 4/ = 2e.
Equivalently, — / = / — e, so (5) becomes
ff KdM = 2x(y  e + /) = 2t X (M). 
Because the EulerPoincare' characteristic is a topological invariant,
this theorem shows that total curvature is a topological invariant.
Explicitly we let M and M be geometric surfaces that are merely diffeo
morphicf Then the Gaussian curvatures K and K of M and M can be
quite different — but their total curvatures are identical, for (being diffeo
morphic) M and M have the same EulerPoincare 1 characteristic; hence
ff KdM = 2ir X (M) = 2ttx(M) = ff_K
dM.
We have already seen instances of this theorem. For example, the torus
in Example 2.5 has K = 0, hence total curvature zero. Alternatively, this
same surface acquires from E its usual geometric structure as a torus of
revolution, for which the curvature is variable — but we found in Chapter
VI, Section 7, that its total curvature is also zero. (The diffeomorphism in
this case is just the identity mapping.)
In general it suffices to count handles to find total curvature.
8.9 Corollary If ¥ is a compact orientable surface with h handles
(h = 0, 1, 2, • • • ), then for any geometric structure on M, the total curva
ture is 4x(l — h).
Proof. We have already seen that M has EulerPoincare" characteristic
2  2h. I
j See footnote, page 380.
Sec. 8] THE GAUSSBONNET THEOREM 383
The GaussBonnet theorem (Theorem 8.8 or 8.9) provides a way to attack
some seemingly formidable problems. For example, (1) of Example 2.3
shows that if one single point is removed from a sphere 2, there exists a
geometric structure on the punctured sphere with K = 0. But there can be
no geometric structure on an entire sphere 2 for which K ^ 0, since then
//.
KdL ^ 0,
contradicting 2ttx(2) = 47r. Reversing this argument, we find that a
compact orientable geometric surface with K > must be diffeomorphic to a
sphere. Its total curvature is positive— but in Corollary 8.9, h is a nonnega
tive integer, so it must be zero. Thus the surface has no handles; it is diffeo
morphic to the sphere 2 = 2(0). Further results of this type are given in
the exercises.
The GaussBonnet theorem is proved by cutting M into rectangular
regions, and applying the GaussBonnet formula to each. The scheme works
because all these regions are consistently oriented by an orientation of M
itself, so that the integrals Jk ds on the boundaries of these regions cancel
in pairs. Here in essence is the fundamental idea of algebraic topology;
indeed considerations of this kind led Poincare" to its invention (see Lef
schetz [8]). By applying this scheme to suitable regions in M we can get a
more general form of the GaussBonnet theorem (Exercise 8). A corollary
(Exercise 11) shows how to extend Theorem 8.4 from rectangles to ar
bitrary polygons. To see how the notion of boundary generalizes in such situ
ations we shall give a direct (and logically unnecessary) proof of Exercise
11 in the special case of a triangle, that is, the onetoone regular image A
of an ordinary triangle T in E 2 (Fig. 7.47).
Corollary 8.10 If A is a triangle in a geometric surface M, then
// KdM + f K g ds = 2tt — (e x + e 2 + e 3 ) = (n + n + n)  v.
(We explain this notation in the course of the proof.)
FIG. 7.47
3*4 RIEMANNIAN GEOMETRY [Chap. VII
y
FIG. 7.48
Proof. Let dM be an arbitrary area form on the region A.
We get a rectangular decomposition of A = y(T) as follows. Cut T
into three quadrilaterals, as indicated in Fig. 7.48; then changes of varia
bles in y will exhibit their images as rectangular regions xi(fli), x 2 (R 2 ),
x 3 (R 3 ) constituting a rectangular decomposition of A. As usual we arrange
for each x* to be positively oriented. Thus, by the GaussBonnet formula
with interior angles, the total curvature of A is
ff K dM = £ ff K dM =  j^ f K.d8br + g
VJA »l JJ %i i=l J dXi
where d is the sum of all interior angles.
Of the twelve edges in d Xl , dx 2 , and dx 3 , the six interior ones cancel in
pairs (at least Jk s ds does, on them). The remaining six combine in pairs
to give the curves a lt a 2 , <x 3 (Fig. 7.47) constituting the boundary dA of the
oriented triangle A. Hence
2 / K dS= Kg dS = Kg dS + / Kg dS + Kg
il J dxi JdA J ai J a2 J az
In the sum 0, the interior angles n, t 2 , t 3 at pi, ps, p 3 are those of the
triangle A itself. The others, which occur at the artificially introduced
vertices, evidently add up to 5tt; thus we find
KdM + J K g dS = U + la + i,)  7T.
•'•'A JdA
We are adapting to the triangle the definitions in 8.3; thus i, + e, = ic
gives the exterior angle formula. 
If the edge curves of the triangle are geodesies, then of course the geo
desic curvature term vanishes. In particular, for a geodesic triangle in a
surface with constant curvature K, this result reduces to
ll + t2 + t3 = 7T + KA
where A is the area of the triangle. Thus the wellknown theorem of plane
geometry that the sum of the interior angles of a triangle is ir depends on
Sec. 8]
THE GAUSSBONNET THEOREM
385
the fact that E 2 is flat. Examples show easily enough how a geodesic tri
angle can manage to have n + i 2 + i 3 larger than jona sphere (K > 0)
and smaller than x on a hyperbolic plane (K < 0) (Fig. 7.49).
EXERCISES
1. Find the total Gaussian curvature of:
(a) An ellipsoid.
(b) The surface in Fig. 4.10.
(c) M : x 2 + y* + z« = 1.
2. Prove that for a compact orientable geometric surface M :
K > =* M is diffeomorphic to a sphere (text)
K = => M is diffeomorphic to a torus
i£ < ^ M is a sphere with /i ^ 2 handles
3. (a) Let M be a compact orientable geometric surface with h handles.
Prove that there exists a point p of M at which
K(p) > if h = 0,
K(p) =0 if h = 1,
#(p) < if A ^ 2.
(b) If M is a compact orientable surface in E 3 which is not diffeo
morphic to a sphere, show that there is a point p of M at which
#(p) < (compare Theorem 3.5 of Chapter VI).
386 RIEMANNIAN GEOMETRY [Chap. VII
4. (a) For a regular curve segment a: [a, b] — » M, show that the
total geodesic curvature /« k ds is
/
dt.
(Hint: Exercise 9 of Section 4.)
(b) Let x be a (positively oriented) orthogonal patch in M. Deduce
the following formulas for total geodesic curvatures of parameter
curves :
(Note that \E V = — x uu *x v and %G U = — x rc «x„, and if M is in E ,
either intrinsic or Euclidean derivatives give the same results.)
5. Let x: R —> 2 be the geographical patch (Example 2.2 of Chapter IV)
restricted to the rectangle R: ^ w, v ^ 7r/4. Compute separately each
term in the GaussBonnet formula for x.
6. If F: M — * N is a mapping of compact oriented surfaces, the degree
d F of F is the algebraic area of F(M) divided by the area of N. Thus
d F represents the total algebraic number of times F wraps M around
N. If M is a compact oriented surface in E 3 , prove the Hopf theorem:
The degree d G of the Gauss mapping is the integer x(M)/2. (It can
be shown that degree is always an integer.)
An oriented polygonal region (P in a surface M is an oriented region that
has a rectangular decomposition x x , • • • , x* which we always arrange to be
positively oriented. Then the boundary d(P of (P is the formal sum of those
edge curves appearing in exactly one of the boundaries dxi, • • • , dx k . We
exclude the situation shown in Fig. 7.50, so d(P consists of simple closed
(broken) curves. These definitions are such that if a is one of the edges
in d(P then J (a) always points into the region (P. (This rigorizes the rough
rule: "Travel around the boundary keeping the region always on the left.")
FIG. 7.50
Sec. 8]
THE GAUSSBONNET THEOREM
387
FIG. 7.51
(a) If is a 1form on an oriented polygonal region (P, prove the gen
eralized Stokes' theorem
// d<t>== I
J J<9 J d<5>
0.
(If d(P = 2 on, then f d(? <j> means S f ai <j>.) (Hint: Lemma 6.6 of
Chapter IV produces some cancellation in pairs, as in the proof
of Theorem 8.8.)
(b) Deduce that if <f> is any 1form on a compact oriented surface M,
then jj M d4> = 0.
(c) Two different (positively oriented) rectangular decompositions
of the same region (P produce technically different boundaries of
(P; however, both occupy the same set of points. Prove that for
any 1form on (P, f d6 > <f> is the same for both.
(Generalized GaussBonnet theorem). If (P is an oriented polygonal
region in a geometric surface, prove that
ft KdM + f K g ds + Y, e i = 2tx(<P)
where 2 e> is the sum of the exterior angles of (P as defined in Definition
8.3 for the special case of a rectangular region (Fig. 7.51).
(Hint: Refine the proof of Theorem 8.8, classifying edges and vertices
into those in d(P and those in the interior of (P. Note that on each simple
closed boundary curve, the number of edges is the same as the number
of vertices.)
Prove that the following properties of a compact orientable surface M
are equivalent:
388 RIEMANNIAN GEOMETRY
[Chap. VII
FIG. 7.52
(a) There is a nonvanishing tangent vector field on M
(b) X (M) = 0.
(c) M is diffeomorphic to a torus.
(Hint: For (a) => (b), introduce a geometric structure and use Exer
cise 7.) Properties (a) and (b) are, in fact, equivalent for any compact
manifold.
10. (a) If a region (R has a rectangular decomposition, derive a triangular
decomposition, and show that v  e + / is the same for both.
(b) Do the same with "rectangular" and "triangular" reversed.
The notion of simple region (Ex. 12 of VI.7) may be extended by allow
ing the mapping F: D * M to be nondifferentiable (but still continuous)
at n points on the circle u 2 + v 2 = 1. This permits n corners to appear in
the boundary d(P of (P = F(D). We call (P in this case an npolygon (n
^ 0) The EulerPoincare characteristic of an npolygon is +1, since
for a triangular decomposition as in Fig. 7.52, we have v  1 = f = e /2
11. If (Pis an oriented geodesic npolygon (the edges are geodesies) in
a geometric surface, show that
IL
K dM = 2t  £ e k = (2  n ) T + £ t .
yi
where e y and i, are the exterior and interior angles of (P.
12. (Continuation), (a) If (P is a geodesic npolygon in the plane, prove
that n ^ 3 and that the sum of the exterior angles of (P is 2x.
(b) If M is a surface with constant Gaussian curvature K ^ 0, show
that the area of a geodesic polygon is determined by its exterior
or interior angles.
(c) In the sphere 2 of radius r, find a geodesic 3polygon (P whose
interior angles are each 3tt/2. What is the area of (P?
13. (a) In a surface M with K ^ prove that there exist no geodesic
npolygons with n ^ 2. Thus, in particular, two geodesies in M
cannot meet to form the boundary of a simple polygonal region.
Sec. 9] SUMMARY 389
FIG. 7.53
(b) On a sphere 2, for which values of n ^ do there exist geo
desic wpolygons? (Do not count "removable vertices" — those
with exterior angle 0. )
14. In the hyperbolic plane, let (P B (n ^ 3) be a "geodesic npolygon"
whose vertices are on the rim u 2 J v = 4 of H, hence are not actually
in H (Fig. 7.53). Find the area of (P n .
15. (Hopf Umlaufsatz) . If /3 is a simple closed curve in E 2 , then the total
geodesic curvature of is ±2x. Thus the unit tangent T of /3 turns
through one full circle in traversing /?. Prove this result assuming
/3 is the boundary curve of a simple region S. (Hint: S is a 0polygon
as defined in the paragraph preceding Ex. 11.)
The assumption above is always true, but its proof requires rather deep
topological methods.
9 Summary
A geometric surface — that is, a 2dimensional Riemannian manifold —
generalizes the Euclidean plane by replacing E 2 by any surface and re
placing the dot product on tangent vectors by arbitrary inner products.
In the resulting Riemannian geometry, the length of a curve is defined as
before, and gives a notion of intrinsic distance directly generalizing the
familiar Euclidean distance in the plane. The acceleration of a curve is also
a geometric notion, but it is not so obvious how an inner product on tangent
vectors can lead to a measurement of the turning of a curve. For 70 or 80
years after Riemann this was accomplished by rather complicated formulas
in terms of coordinate patches (4.2 is a sample). In the Cartan approach,
the inner products serve to define the notion of frame field, and the rate
of rotation of a frame field is expressed by its connection form. The con
nection equation V y #i = o} n (V)E 2 then defines the covariant derivative—
with acceleration as a special case.
In Riemannian geometry as in Euclidean geometry, geodesies are the
390 RIEMANNIAN GEOMETRY [Chap. VII
curves with acceleration zero. Geodesies are not only straightest curves,
however; they are also shortest curves in the sense discussed in Sections
5 and 6. The simple Euclidean rule that "a straight line is the shortest
distance between two points" is no preparation for the new and subtle
behavior of geodesies on an arbitrary geometric surface — or even a surface
as simple as a sphere or a cylinder. Some idea of how far the analysis of
geodesies can lead may be found in Milnor [7].
It is by now hardly necessary to say that the Gaussian curvature K of a
geometric surface M is its most important geometric property, for we have
seen that sooner or later curvature enters into almost every geometrical
investigation. Indeed K could be defined, for example, in terms of parallel
vector fields (holonomy) or radial geodesies (the Jacobi equation) or
polar circles. (For a surface in E 3 we used the shape operator, and could
have used the Gauss mapping.) In the Cartan approach, however, curva
ture is defined by the structural equation duu. = —K6 X * 2 , which pre
sents K (in a sense discussed earlier) as the common "second derivative"
of all frame fields on M . And it is this definition that leads most directly
to the central result of twodimensional Riemannian geometry, the Gauss
Bonnet theorem. Leaving aside trigonometric consequences such as Corol
lary 8.10, the content of the theorem is that curvature determines topol
ogy, at least in the compact orientable case.
Generally speaking, the results of this chapter are valid for Riemannian
manifolds of arbitrary dimension n, and in most instances the definitions
and proofs need scarcely any readjustment. Dimension 2 has simplified
certain consistency proofs such as Theorems 2.1 and 3.2, but these can
be avoided entirely by more advanced methods. As one might expect, it is
the GaussBonnet theorem whose generalization offers the most difficulty
(see Hicks [5]), and in higher (even) dimensions the curvature of M in
fluences but does not control the topological configuration of M.
Bibliography
1. H. Flanders, "Differential Forms: With Applications to the Physical Sciences."
Academic Press, New York, 1963.
2. G. Birkhoff and S. MacLane, "A Survey of Modern Algebra." Macmillan, New
York, 1953.
3. T. J. Willmore, "An Introduction to Differential Geometry." Oxford Univ. Press,
London and New York, 1959.
4. R. Courant and H. Robbins, "What is Mathematics?" Oxford Univ. Press, London
and New York, 1941.
5. N. J. Hicks, "Notes on Differential Geometry." Van Nostrand, Princeton, New
Jersey, 1965.
6. D. J. Struik, "Lectures on Classical Differential Geometry." AddisonWesley,
Reading, Massachusetts, 1961.
7. J. W. Milnor, "Morse Theory." Princeton Univ. Press, Princeton, New Jersey,
1963.
8. S. Lefschetz, "Introduction to Topology." Princeton Univ. Press, Princeton, New
Jersey, 1949.
The books of Willmore and Struik are at about the same level of difficulty as this
one. Hicks' book begins at a level comparable with our Chapter VII and gives a very
concise exposition of multidimensional Riemannian geometry; its bibliography lists a
number of more detailed works on the subject.
Answers to Odd
Numbered Exercises
(These answers are not complete; and in some cases
where a proof is required, we give only a hint.)
Chapter
Section 1
1. (a) x*y* sin 2 z, (c) 2x*y cos z
3. (b) 2x<* cos (e*), h = x 2 + y 2 + z 2
Section 2
1. (a) 6C7i(p) + U»(p)  9C7 3 (p)
3. (a) V = (2z*/7)Ui  (xy/7)U 3
(c) V = xUi + 2yU 2 + xy*U 3
5. (b) Use Cramer's rule.
Section 3
1. (a) 0, (b) 72 7 , (c) 2e 2
3. (a) y\ (c) yz*(y 2 z  Sx 2 ), (e) 2x(y*  3z 5 )
5. Use Ex. 4.
Section 4
1, «'(x/4) = (2,0,V2)p, where p = (1,1,V2)
3. /3(s) = (2(1  s 2 ), 2s Vl  s 2 , 2s)
5. The lines meet at (11,7,3)
7. v„ = (1,0,1),
9.Ata(0); «> (2,2*,*)
Section 5
1. (a) 4, (b) 4, (c) 2
5. (b) (xdy  ydx)/(x 2 + 2/ 2 )
393
394 ANSWERS TO EXERCISES
7. (a) dx — dz, (b) not a 1form, (c) zdx + x dy, (d) 2(xdx + ydy), (e) 0, (f) not
a 1form
9. ± (0, 1, f)
11. (a) Use the Taylor approximation of the function t — >/(p + t\)
(b) Exact: —0.420, approximate: — \
Section 6
1. (a) <f> A $ = yz cos z dx dy — sin z dx dz — cos z dy dz
(b) d^ = —zdxdy — ydxdz. Note that d(dz) = d(ldz) = Oby 1.6.3.
7. Apply this definition to the formula following 1.6.3.
9. Assuming the formula, set / = y, g — x.
Section 7
1. (a) (0,0), (b) (3,1), (3,1), (c) (0,0), (1,0)
5. (a) (2,0,3) at (0,0,0), (b) (2,2,3) at (0,2,tt)
7. GF = (flritfu/O.fcC/i,/.))
11. (a) F* = (», ue~°), (b) F 1 = (m 1 ' 3 , t> + « 1/3 ), (c) F" 1 = ((9  u  2»)/2, 5 
tt — i>). F is a diffeomorphism for (a) and (c) only, since in (b), F~ l is not differ
entiable (when u = 0).
Chapter II
Section 1
1. (a) 4, (b) (6,2,2), (c) (1,2,1)/V6, (1,0,3)/Vl0, (d) 2^11, (e) 2/V15
5. If v X w = 0, then u»v X w = for aK u; use Ex. 4.
7. V2 = v — (v»u)u
Section 2
3. 0(8) = (VI + s 2 /2, *A/2, sinh" 1 (*A/2))
5. If /Si is based at ti{i = 1,2), then so is plus or minus the arc length of a from t\ to
fe.
9. (b) The condition is certainly necessary; to see that it is sufficient, show that a
unit speed reparametrization of a. has acceleration zero.
11. (b) L(a) > / a'u dt = Y, r 1 Uidt = (q  p)»u = d(p,q)
•>« J« *
Section 3
1. K = l, T = 0, B = (,0,), center (0,1,0), radius 1.
7. (a) 1 =  a(h)'  =  a'(h)h'  =  h' , hence A = ±1
(b) Let e = ±1, then a = a(h) implies T = a' (h)h' = eT(h); hence
«iV = k(/i) N(h), and so on.
9. The orthogonal projection on the N B plane (the normal plane of /3 at /3(0)) is
s » Ko (sV2)iVo + k t (s 3 /Q)Bo (cusp at s = 0).
ANSWERS TO EXERCISES 395
1 1. B = B implies tN = fN, hence either (1) f = r and N = N, or (2) f = — r and
N N.
Section 4
1. Let/ = t* + 2; then K = r = 2//»; 5 = {t\2t,2)ff
3. (a)JV(O) = (0,1,0), r(0) = f
9. (a) * = x/4, u = (1,0,1) A/2, t(0 = «  (*76), < 2 , « + « 3 /6))
15. (c) The evolute is also a cycloid.
17. «'(*) = (/(*) sin «,/(*) cos t,f(t)g(t))
Section 5
1. (a) 2tfi(p)  t/ 2 (p), (b) C/^p) +2[/ 2 (p)+4 C7 3 ( P )
5. (a) 8tf,(p)  4(t/ 3 )(p)
Section 6
1. Show that V'W = 0, and use III .8.
3. For instance, E 2 = —sin 2t/ 2 + cos zU 3 , and E z = E t X E 2 .
Section 7
1. W12 = 0, W13 = W 2 3 = (d/)/\/2
3. wi2 = — df, con = cos / df, w 2 3 = sin f df
5. By (3) of II.5.4, V v (E/^i) = £ Ff/;]^ + E/iV^.
7. At an arbitrary point p, a(t) = fp is a curve with «' = [ p [ Fi. Show that
II P II Fi\p] = II P .
Chapter III
Section 1
3. (T a )~ l = T_ a , CM = «C, hence F" 1 = (7\,C)i = C 1 ^ = TVci^C" 1 .
5. (b) Using Ex. 3, we find F~i(p) = (5y/2, 5, 4V2)
Section 3
1 . If F and G have orthogonal parts A and B, then by III. 1 .2 , sgn (F(?) = det (AB)
= det ^det B.
5. C is a rotation, through angle ir/2, around the axis given by a.
7. For E 1 : F(s) = es + s ; for E 2 : F =
m /cos # « sin #\
:F=TC, where C = ( J.
\sin # e cos #/
Section 4
1. (b) By definition, p(s) is the point canonically corresponding to T(s), hence by
III.2.1, 0(0) corresponds to ^(T 7 ), the unit tangent of F(0).
5. For a tangent vector v at p, F* (V V W) = W(F(p) + *C(v))'(0) = V ,+ M W.
396 ANSWERS TO EXERCISES
Section 5
1. P = T p (C(a)), where C( Ui ) = e,
3. Consequence of III.5.7
5. If t is not identically zero, assume r (0) ^ and examine the proof of 5.3.
7. Let i? = TC, where T is translation by (0,0,bs /c) and
(cos (s /c) e sin (s /c) 0\
sin (s /c) e cos (s /c) J
J
where e = ±1. Then F(/8) = /3(«s + s„)
9. a(s) = (/ cos <p(s) ds, / sin *>(s) ds), where *>(«) = / k(s) ds
11. Use Ex. 9
Chapter IV
Section 1
1. (a) The vertex, 0, (b) all points on the circle x* + y* = 1, (c) all points on the z
axis
5. (b) c ^ 1
9. Use Ex. 7
1 1. q is in F(M) if and only if F~ l (q) is in M, that is, g(F~Hq)) = c. Use the hint
to apply IV.1.4.
Section 2
1. (c) x(u,v) = (u,v,v? + v 2 ) is one possibility; a parametrization derived from
IV.2.5 will omit a point of the surface.
5. x u X x v = vd' X 5
7. (b) Straight lines (rulings) and helices, (c) M: x sin (z/b) = y cos (2/6)
9. x(w,y) = (cos u — v sin w, sin w + v cos w, i>)
13. (a) If g' is never 0, reparametrize the profile curve to obtain u — >■ (u,f(u),0),
and use IV.2.5.
Section 3
1. (a) r 2 cos 2 v, (b) r 2 (l — 2 cos 2 v cos w sin u)
3. (a) m and v are the Euclidean coordinate functions of x x y
(b) Express y = x(u,v) in terms of Euclidean coordinates, and differentiate.
5. (a) M is given by g = z  /(ar,y) = 0, with Vj = (/„/ y ,l), and v is tangent
to M at p if and only if vV^(p) = 0.
7. Vg = (— 2/,— a;,l) is a normal vector field; V is a tangent vector field if and only
if V*Vg = 0, for example, V = (x,0,z).
9. (a) T P (M) consists of all points r such that (r — p). z = 0; hence v p is in T P (M)
(that is, vz = 0) if and only if p + v is in T P (M).
11. (a) 2tt
ANSWERS TO EXERCISES 397
Section 4
= (d/ a + /dtf>) (x„ , xj
5. If a is a curve with initial velocity v at p, then
yp[g(f)] = (ftf«)'(0) = </'(/«) (0)(/«)'(0) = 0'(/(p) )▼*[/]•
7. On the overlap of ^f and 11 ,, d/»  df,  d(/<  //) =
9. (b) d«t(x„) = x„M = — — = 1.
du du
Section 5
1. If x: D » M is a patch, then F(x) : D ► JV is (by 3.2) a differentiable mapping.
Hence y 1 Fx is differentiable for any patch y in AT.
3. If x and y are patches in M and N, respectively, then y _1 Fx = (y 1 y) (x 1 x) is
differentiable, being a composition of differentiable functions.
7. (b) x*(»») = r 3 sin 2 v cos v du dv
11. Only (a) is not a diffeomorphism.
13. (b) F*(ax u + bx v ) = ay„ + by v implies linearity.
Section 6
7. (a) 2irw, (b) 2jrn
13. (b) Show f a <\> = Jx d<t> for a suitable x.
15. Use simple connectedness to prove the formula given in the hint — see Fig. 4.46.
Section 7
1. (a) Connected, not compact, (c) connected and compact, (e) connected, not
compact.
3. If v is nonvanishing on N, show that F*v is nonvanishing on M .
5. (a) If Z is a nonvanishing normal, then let ±U = ±Z/  Z \\. If V is an arbitrary
unit normal, write V = (V'U)U and use Ex. 4(b).
(b) The image x(D) of a patch with D connected.
9. (c) Use Ex. 7
Section 8
1. Modify the proof given for the Mobius band in IV.7.
9. (x X y) _1 (x X y) = (x _1 x) X (y _1 y), a differentiable function.
398 ANSWERS TO EXERCISES
Chapter V
Section 1
1 . Use Method 1 in text.
3. (a) 2, (c) 1
Section 2
1. (b) If ei, e 2 = (ui ± u 2 )/V2, then £( ei ) = ei and S(e 2 ) = e 2
Section 3
5. (b) An ellipse on one side and no points on the other; the two branches of a
hyperbola (asymptotes the two lines in (a)); two parallel lines on one side, and
no points on the other.
7. (a) If a is a curve in M with initial velocity v at p, then F*(v) = F(a)'(0) =
(a + eU a )'(0) = v  eS(v) at F(p).
Section 4
5. K = 36r 2 /(l + 9r 4 )*; not minimal
7. Compute speed from a' = a/x M + a 2 'x v
13. p = x(u,v) is umbilic if and only if S(x u ) = Ax„ and S(x v ) = Ax, at (u,v). Dot
with x„ and x v .
15. (a) None, since K < 0, (b) origin (a planar point),
/ b a 2 _ ^ 2 \
(c) I 0, ±  Va 2  6 2 , —  — ) if a > b
Section 5
3. A meridian a lies in a plane orthogonal to M along «, hence a" is tangent to this
plane, and (with constant speed parametrization) orthogonal to «'; thus a" is
orthogonal to M.
7. S(T) = —U', hence by orthonormal expansion, U' = — S(T)»T T — S(T)»V V.
Continue as in the proof of the Frenet formulas.
15. On the ruling through <r(u), the formula for K in Ex. 14 shows that either K is
identically zero, or K < has minimum value —\/p(u) 2 at a(u) and rises sym
metrically toward zero as v — * ± °°
17. (a) For (w,0,0) + v(0,l,u): the x axis, with p(u) = 1 + u 2 . (b) Use Ex. 15. For
fixed u, K =  (1 + m 2 + ^)2 is a minimum when v = 0.
19. (c) x = a + v8 is noncylindrical, and we can assume that a is a (unit speed)
striction curve. But a'»S X 5' = (since K = 0) and a'S' = (striction), hence
T = a! and 5 are collinear.
Section 6
T. K = (1  z'Xl + x* exp (a; 2 )) 2
3. Use the results of V.2. Note that meridians are normal sections.
5. M has parametrization x(u,v) = (u cos v, u sin v, /(«)).
ANSWERS TO EXERCISES 399
7. With the usual parametrization, an argument like that for V.6.2 reduces to the
extreme cases: g' always zero, g' never zero. In the former, M is part of a plane
(special case of cone).
9. (c) If y = f(x) has unit speed parametrization (g,h), where h(u) = ce~ ulc , show
that / (not h !) satisfies the differential equation in VI .6.6.
Chapter VI
Section 1
1 . (a) a" = ui2 (T)E 2 + a>i3(T)E 3 , hence a" is normal to M if and only if ui 2 (T) = 0.
5. If the cylindrical frame field is restricted to M and the indices 1 and 3 are reversed,
we obtain the frame field in (1) of VI.1.3. By the computations in II.7,
0>12 = «13 = 0, 0>23 = — d&
Section 2
3. (a) yp = *CEi)«i + ^(E 2 )e 2 = Mi + fad 2 .
(b) f = hio) 23 — hvmz
Section 3
I . If K = H = 0, then fafa = h + k 2 = 0. Thus h = fa = 0, and S = 0.
3. Assume fa 9^ fa, and use the Hilbert Lemma (3.6) to get a contradiction.
5. In the case fa ^ fa, use VI.2.7 to show that, say, fa = 0. By Ex. 2 the k\ princi
pal curves are straight lines. Show that the fa principal curves are circles, and that
the (&i) straight lines are parallel in E 3 .
Section 4
1. (d) => (b): if u is an arbitrary tangent vector at p, then u = av + bw, hence
 F*u  2 = a 2  F*v  2 + 2ab F*vF*w + 6 2  F*w  2 =
a 2  v  2 + 2ab vw + fc 2 \\ w  2 =  u  2
3. If a is a curve segment from (—1,0,0) to (1,0,0) with length 2, then by the exercise
mentioned, a parametrizes a straight line segment — impossible, since a must
remain in M.
5. (a) Define F( a (u) + vT a (u)) = 0(u) + vT„{u)
(b) Choose fi in E 2 with the same curvature function.
7. (a) Criterion (a) becomes F*{\)*F*{\f) = X 2 (p)v»w; criterion (c) becomes
n( ei ).F*( e/ ) = x 2 ( P )5 iy .
II. Write F(\(u,v)) = x(a(u), b(v)) for suitable parametrizations.
13. For y, show that the conditions E = G and F = are equivalent to g' = cos g,
which has solution g(v) = 2 tan 1 (e v ) — (ir/2) such that g(0) = 0. Use Ex. 7.
15. F(x(u,v)) = (f(u) cos v,f(u) sin v), where x is a canonical parametrization and
f(u) = exp ( /? (dt/h(t)).
400 ANSWERS TO EXERCISES
Section 5
1. If a' = Ex along a , then F( a )' = F* (Ex) = Ex along F( a ). Use VI.5.3.
3. There is no local isometry of the saddle surface M (1 < K < 0) onto a catenoid
with — 1 < K < 0, since K has an isolated minimum point (at 0) while K takes
on each of its values on entire circles. (Many other examples are possible.)
5. (b) Follows from VI.4.5, since for x t we compute E t = cosh 2 u = G h and F t =
(independent of t).
(d) For M t :U t = (s,c,S)/C, so the Euclidean coordinates of U t are independent
of t.
Section 6
1. (b) Ox = Vl + u 2 du, 02 = u dv, «i2 = dv/Vl + u 2 , K = 1/(1 + u 2 ) 2 .
3. «12 = — t? u rft<
Section 7
3. (a) yl = (2tt/3){(1 + c 2 ) 3 ' 2  1}, (b) <*>
5. (a) Use a canonical parametrization; then x*(KdM) = (h" /h)(h du dv) =
— h" du dv. Recall that h' = sin <p.
(c) for the bugle surface, lim <p a = —1, lim <p b = 0.
a»0 6*oo
7. Use Ex. 5. At the rims of all these surfaces, h' — sin*?— ► ±1. For K = 1/c 2 (V.6.5):
in Case (2), TC = <Lwa/c, A = 4irac; in Case (3), TC = 4tt, A = 4ttc 2 . For if =
1/c 1 (Ex. 9 of V.6) : M a has TC = 2x(a  c)/c, M b has TC = 4,r.
9. (a) Use Ex. 14 of V.5.
L
pi
dv = 2.
( p 2 + ^)3/2
11. Define x(u,v) = F((l  u) cos i;, (1  u) sin »), on #: < u < 1, < » < 2tt.
13. With outward normal, H = — 1/r, and h = r.
1 5. Dividing if necessary by the lengths of V and W, we can assume they are unit
vector fields. Thus V, U X V is a tangent frame field on a. Orthonormal expan
sion shows that if / = W*V and g = W*U X V, then/ 2 f g 2 = 1.
17. (a) Use  ^v X F*w  =  J   v X w 
19. (a) F carries positively oriented pavings of M topositively oriented pavings of N.
(b) If F: M — ► M is an isometry, orient M and M so that F preserves orientation
(see remark following Ex. 4). Then
JJ KdM = ff F*(K dM) = ff K(F)F*(dM) = ff K dM
Section8
1 . If A^ is isometric to a sphere 2 of radius r, show that N is compact and has K =
1/r 2 . Thus,vby Liebmann's theorem, N is also a sphere of radius r, so a translation
will show that N is congruent to S.
3. Except for geodesies, all follow immediately from preservation of shape operators.
Only geodesies and Gaussian curvature need be preserved by arbitrary isometries
(Ex. 1 of VI.5, and theorema egregium).
ANSWERS TO EXERCISES 401
5. (b) That given by  U.
9. Deduce from theorema egregium that a Euclidean symmetry F of M must leave
the origin fixed, hence F is orthogonal. Consider its effect on the natural frame
at 0.
Chapter VII
Section 1
1. (b) <x'°a' = a'*a'/g 2 (a) is speed squared.
3. (b) Let Ei and Ez be the vector fields on x(D) determined by x u /y/E and V /  V \\,
where V = x v — (F/E)x u (GramSchmidt process).
5. (a) Show that the definition J(Ei) = E 2 ,J(E 2 ) = Ei is independent of the choice
of positively oriented frame field. (Any two have the same orientation in the sense
defined just preceding VII.1.4.)
(b) For/ 2 = /, show J(J(Ei)) E t .
7. (a) F*£7, = f u Ui + gJJ* , F*t/ 2 = f t Ui + g v U, (Use Ex. 6).
(b) The complex derivative of F = / + ig is dF/dz = /„ + ig« ■ Thus the magnitude
 dF/dz  is (f u 2 + g u 2 ) 112 , which may be rewritten in various ways using the Cau
chyRiemann equations.
Section 2
1. 0i = du/v, &2 = dv/v
3. A = 7rr 2 /f 1  — J , (E, F, G computed as in Example 4.11); A(H) = <».
5. E = G = 1 and F = 0, so A = 4ir 2 . We can redefine the geometric structure to make
E and G any positive numbers.
7. Show that the parametrization x in Ex. 5 of VI.5 is an isometry.
Section 3
1. Note that E, = vUi restricted to a is r sint [/, , thus a' = —E x + cot (0^2 . Since
wis = du/v evaluated on a' is — 1, we derive from the covariant derivative for
mula that a" = cot (t)Ei  cot 2 (t)E 2 .
3. In the proof referred to, note that u> l3 (V)E 3 = V v EfEiE 3 = S(V)'EiE 3 .
5. (a) The proof of VII.3.6 shows that the holonomy angle of a is
«6
— / Wl2(a') dt = — I C012
Use Stokes' theorem, recalling that dwvi = —K dM (first structural equation).
7. Y' = /'_Ei + fw2i{oc')E 2 , hence F„(F') = f'E, + }un{a')Et . F.Y = Y = /£, ,
hence F' = /'F, +/»ii(F,a')^i • Use VI.5.3.
11. H 7 has constant length c; on any orientable region, show that the frame field
Ei = H'/c, F 2 = /(Fi) has connection form zero, hence curvature zero.
402 ANSWERS TO EXERCISES
Section 4
1. Since a" = 0, a{h)" = a '(h)h"
3. Apply Ex. 6 of Section 3.
5. All Euclidean circles through the north pole, since these correspond under sten
ographic projection to straight lines in the plane.
9. Using the equations for a.' and a." given in the text, we have J (a') = J (vT) = vN
hence a" •/(«') = Kg v 3 .
11. (a) For u <u < Ul  t> a/ = ^° _ & > A t >
y/EG ~
(b) // the meeting takes place, then <*' and 0' are collinear, hence by VII.4.2, «
and are equal (but for parametrization)— an impossibility.
13. (a) for the usual parametrization of a surface of revolution, G = h 2 (h distance
to t'he axis of revolution), and the wparameter curves are the meridians, hence
the slant c is h sin <p.
(b) Follows from Exs. 11 and 12, since such a parallel is a barrier curve.
15. Meridians obviously approach the rim (in one direction). Use Ex. 13 to show
that any other geodesic meets one barrier curve and approaches the rim in both
directions.
17.  sin <p j < 1/cosh m .
19. (a) Use VI.2.4, (b) Kl = 0, k 2 = h'/h.
Section 5
1. p(0, p) = tanh _1 ( p /2), Euclidean norm.
3. The geodesies are helices, and y(u, v) = y U e 1+ ve 2 (l) = (r cos u/r, r sin u/r, v).
The largest normal neighborhood is r, rr ; y is regular for any «, but the oneto
one condition fails for e > irr.
5. (a) If q is the r, e , then by 5.5, p(p, q) < e . If q is not in r, t a curve from p to q
meets Cs for every S < e.
(b) If p ^ q, it follows from the Hausdorff axiom that there is a normal neighbor
hood of p which does not contain q, hence p(p, q) > 0.
7. The length L of a from to r + e is the same as that of the broken curve: from
to r, then a. from r to r + *. Hence L > p(a(0), a (r + e)) by the remark preced
ing 5.8.
9. D: u 2 + v 2 < 1 with Euclidean geometric structure.
11. (a) There is only one geodesic segment (a meridian) from p to any other point.
Section 6
5. (a) x u (0, v) = X(v), and since x(0, v) = 7xoo(0) = p(v), we have x„(0, v) = 0'(v).
Thus EG  F 2 is nonzero when u = 0, and by continuity for  u \ small.
(b) (iii) a base curve, X = S.
7. (a) The yparameter curves are meridians of longitude.
(b) Since K = 0, the Jacobi equation becomes (VG)«« = 0, hence y/G is linear
in u, and it follows that G(u, v) = 1 — Kg (v)u.
ANSWERS TO EXERCISES 403
Section 7
1. If N is geodesically complete, fix a point p of M ; then there is a geodesic segment
y w from F(p) to an arbitrary point q of AT. But for v such that F*(v) = w, we
have F(y v ) = y w , hence q is in F(M).
3. If p 5* q in M, then any geodesic segment a from p to q has nonzero speed. Hence
F(<r) is a nonconstant geodesic from F(p) to F(q), and it follows that F(p) ^
F(q), even if "two" is interpreted as "two distinct" in the uniqueness property
of TV.'
5. Variant of VII.7.3.
7. To show that a surface M is not frame homogeneous, it suffices (by the argument
given in the text for the cylinder) to show that M has some geodesies which are
onetoone and some which are not. Note that x in VII.2.5 is a local isometry.
9. If L is the arc length of the ellipse, show that F(x(u, v)) = x(u + (L/4), i>) is
an isometry of M which does not preserve Euclidean distance.
11. By Ex. 8 (b), F is the restriction to 2 of a linear (orthogonal) transformation,
hence F(p) = F(p). Thus we_can define F{p, p) = {F(p), F(p)}, and
deduce the framehomogeneity of 2 from that of 2.
13. The function is a homomorphism, so it suffices to prove it is onetoone. But in
the proof of VI.8.3, if M is not planar (so S ^ 0), then F is unique.
Section 8
1 . In (a) and (c) the surface is diffeomorphic to a sphere, so TC = 4*. In (b) there
are four handles, so TC = —12w.
3. (a) For h = 1 : if K were never zero, then by an earlier exercise, either K > or
K < 0; both are impossible, since by VII.8.9, //m K dM = 0.
(b) By VI.3.5, K is somewhere positive. But M has at least one handle, so
//.
K dM < 0;
M
hence K is somewhere negative.
5. J"/* KdM =  $to.K ds = x/4\/2
7. (a) If xi , • • • , Xjt is a positively oriented decomposition, then by Stokes theorem
(IV.6.5) we have /<p d<t> = 2 JUi d<t> = 2 /<*, 4> But for edges not in d(P, IV .6.6 pro
duces cancellation by pairs.
(c) If Xi and y; are two positively oriented decompositions, then by the remark
following VI.7.5, we have
Jdyj JJ yj JJ *i J 3*i
9 ( a ) => (b) : Any geometric structure onM (such exist) can be modified so that the
nonvanishing vector field Fi has unit length; then Fi , J(E X ) is a frame field.
Its connection form is defined on the entire surface, so 2ttx(M) = JJm K dM =
— Jf M do>i2 = 0.
1 1 . Consequence of Ex. 8.
13. By the uniqueness of geodesies, interior angles of a geodesic polygon cannot be
±7r (that is, there are no cusps), so — ■* < e < ir. Use Ex. 11.
(b) all n 5^ 1 (for n = 0, a great circle).
INDEX
Acceleration, 54, 68
of a curve in a surface, 196, 324(2&r. 3)
intrinsic, 320
Adapted frame field, 246251
Algebraic area, 289290, 386 (Ex. 6)
Allumbilic surface, 257259
Alternation rule, 27, 47, 153
Angle, 44
coordinate, 210
exterior, 374
interior, 374
oriented, 291
Angle function, 50(Ex. 12), 295(£c. 15)
Antipodal mapping, 165 (Ex. 5)
Antipodal points, 182
Arc length, 5152, 220(Ex. 7)
function, 51
Area, 280286
Area form, 283284, 292, 309(#x. 4)
Areapreserving mapping, 295(Ex. 16)
Associated frame field, 276277
Asymptotic curves, 226227, 230
Asymptotic directions, 225227
Attitude matrix, 46, 88
B
Basis formulas, 251
Bending, 268, 275(Ex. 5)
Binormal, 57, 66, 69
Boundary
of a region, 386
of a 2segment, 170
Bracket operation, 81 (Ex. 7), 195(Ex. 9)
Bugle surface, 241242, 244(Ex. 9), 282
283, 291
Canonical isomorphism, 78, 44, 59
Canonical parametrization, 238
Cartan, E., 42, 92, 96, 303
Cartesian plane, 306
Cartesian product, 187 (Ex. 9)
Catenoid, 236
Gaussian curvature, 236, 239
total, 287288
Gauss mapping, 296(2&c. 20)
local isometry onto, 267268
as minimal surface, 236238
principal curvatures, 236
Center of curvature, M(Ex. 6)
Circle, 59, 62
Clairut parametrization, 330339
Closed surface in E 3 , 181 (Ex. 10), 263n
Codazzi equations, 249, 255
Compact surface, 176177, 259260
Complete surface, see Geodesically com
plete surface
Composite function, 1
Cone, 140(Ex. 5), 2Sl(Ex. 11)
Conformal geometric structure, 305306,
SW(Ex. 1), 312313
Conformal mapping, 268271, 310
Conformal patch, 270(Ex. 7), 279(Ex. 2)
Congruence of curves, 116123
determined by curvature and torsion,
117
Congruence of surfaces, 189, 297303
Conjugate point, 353359
Connected surface, 176
Connection equations
on Euclidean space, 86, 248
on a surface, 248, 318
Connection forms
on Euclidean space, 8590
on a surface, 248, 272, 277, 306
Conoid, 233 (Ex. 20)
Consistent formula for a mapping, 165
(Ex. 10)
Coordinate angle, 210, 22Q(Ex. 8)
Coordinate expression, 143, lQ5(Ex. 6)
Coordinate patch, see Patch
Coordinate system, l58(Ex. 9), 276277
Covariant derivative
Euclidean, 7780, 116CEc. 5), 189190,
321
405
406
INDEX
intrinsic, 318326
on a patch, 212, 324326
Co variant derivative formula, 91 (Ex. 5),
189190, 321
Cross product, 4749, 107, 110
Crosssectional curve, 138
Curvature, see also Gaussian curvature
of a curve in E 2 , 65(Ex. 8), 122123
of a curve in E 3 , 57, 66, 69
Curve, 15
closed, 169
coordinate expression for, 144
oneone, 20
periodic, 20
plane, 61
regular, 20
simple closed, 151ra
in a surface, 144145
unparametrized, 2021
Curve segment, 5Q(Ex. 10), 167
Cylinder, 129, 231 (Ex. 11)
geodesies, 275(Ex. 2), S52(Ex. 8)
parametrization, 137, 141 (Ex. 6)
Cylindrical frame field, 8283
connection forms, 8990
dual 1forms, 9Q(Ex. 4)
Cylindrical helix, 7275
D
Darboux, 81
Degree
of a form, 27, 152
of a mapping, 386 (Ex. 6)
Derivative of Euclidean vector field, 113
Derivative map, 3540
of an isometry, 104
of a mapping of surfaces, 160161, 166
of a patch, 149 (Ex. 4)
Diffeomorphism, 38, 40(Ex. 11), 161
Differentiability, 4, 10, 33, 143, 145146
Differential, 2326
Differential form
closed or exact, 157 (Ex. 2), 173175
on E 2 , 156
on E 3 , 2231
on a surface, 152157
pullback, 163
Direction, 196
Directional derivative, 1115, 149
computation of, 12, 25
Director curve, 140
Distribution parameter, 2Z2(Ex. 14)
Domain, 1
Dot product, 4244, 46, 82, 210
preserved by isometries, 105
Dual 1forms, 9192, 248, 306
Dupin curves, 208(Ex. 5)
E
E, F, G, \4Q(Ex. 2), 210, 3H(Ex. 8)
Edge (curve), 170, 373, 377
Ellipsoid, l42(Ex. 10)
Euclidean symmetries, 303 (Ex. 10)
Gaussian curvature, 217219, 222(Ex
19)
isometry group, 365
umbilics, 223 (Ex. 23)
Elliptic hyperboloid, 142(Ex. 10), 221222
Elliptic paraboloid, 142(Ex. 11), 220(Ex.
6)
Enneper's surface, 221 (Ex. 12)
Euclid, 335336
Euclidean coordinate functions, 9, 15, 24,
33, 53
Euclidean distance, 43, 49 (Ex. 2)
Euclidean geometry, 112, 304305, 308
Euclidean plane, 5, 305, 335336
local isometries, 363364
Euclidean space, 3, 5
Euclidean symmetry, 302303, 365
Euclidean vector field, 147
EulerPoincare" characteristic, 378380
Euler's formula, 201
Evolute, 75(Ex. 15)
Exterior angle, 374
Exterior derivative, 2831, 3132, 154155
Faces, 377
Flat surface, 207, 2Sl(Ex. 11), 263(#r. 2)
Flat torus, 316, S17(Ex. 5), S70(Ex. 7)
imbedding in E 4 , 369
Focal point, 362
Form, see Differential form
Frame, 44
Frame field
adapted, 246
on a curve, 120
on E 3 , 82
natural, 9
INDEX
407
principal, 254
on surface, 251, 306
transferred, 272273
Framehomogeneous surface, 366
Frenet, 81
Frenet apparatus, 63 (Ex. 1), 71
Frenet approximation, 6061, 65 (Ex. 9)
Frenet formulas, 58, 67
Frenet frame field, 57
Function, 12
onetoone, 2
onto, 2
Gauss, 245, 274, 303
GaussBonnet formula, 375377, 383385,
388(Ex. 11)
GaussBonnet theorem, 380383, 387 (Ex.
8)
Gauss equation, 249
Gaussian curvature, 203207, 310312, see
also individual surfaces
differentiability, 206
formulas
explicit, 206, 212, 217, 253, 278
implicit, 205, 252
Gauss mapping and, 289
geodesic curvature and, 339(Ex. 19)
geodesies and, 355358
holonomy and, 325 (Ex. 5)
interval, 275(Ex. 3)
isometric in variance, 273275
in Jacobi equation, 355
polar circles and, 359360
polar discs and, 361 (Ex. 2)
principal curvatures and, 203
shape operator and, 203
sign, 204205
Gauss mapping, 194195, 289291
Geodesically complete surface, 263n,
328329
shortest geodesic segments in, 348
Geodesic curvature, 230(Ex. 7), 329330,
337 (Ex. 9)
total, 372373
Geodesic polar mapping, 340341
Geodesic polar parametrization, 341344
Geodesies, 228232, 326363, see also in
dividual surfaces
closed, 232(Ex. 13)
coordinate formulas, 327, 333
existence and uniqueness, 328
frame fields and, 250(Ex. 1)
lengthminimizing properties, 339359
preserved by (local) isometries, 275
(Ex. 1), 326, 362363
spreading of, 353354
on surfaces of nonpositive curvature,
358, 388(Ex. 13)
Geographical patch, 134^135
Geometric surface, 305, 308
Gradient, 32(Ex. 8), 50(Ex. 11)
as normal vector field, 148, 216
Group, 103
Euclidean, 103
Euclidean symmetry, 302(Ex. 7)
isometry, 365
orthogonal, 104
H
Halmos symbol, 9
Handle, 379, 382
Hausdorff axiom, 186(Ex. 5), 345n
Helicoid, 141 (Ex. 7)
Gauss mapping, 296 (Ex. 20)
local isometries, 267, 275276
patch computations for, 213214
as ruled minimal surface, 227, 233 (Ex.
22)
Helix, 1516, 5859, 119
Hilbert's lemma, 261
Hilbert's theorem, 263
Holonomy, 323325
angle, 323
Gaussian curvature and, 325 (Ex. 5)
Homogeneous surface, 366368, 370
Homotopic to constant, 175
Hopf, 386, 389
Hyperbolic paraboloid, 143 (Ex. 12), 220
(Ex. 6)
Hyperbolic plane, 315316
geodesic completeness, 351 (Ex. 1)
geodesies, 334336
isometries, 371 (Ex. 14)
local isometries, 364
of pseudoradius r, 317 (Ex. 4)
Identity mapping, 99
Image, 1
40(1
INDEX
Image curve, 33
Immersed surface, 187 (Ex. 10), 219, 368
Improper integral, 285286
Induced inner product, 308, 313
Initial velocity, 21 (Ex. 6)
Inner product, 304, 316317
Integral curve, 186187
Integral of function, 286, 292 (Ex. 4)
Integration of differential forms, 167
176, 283297
1 forms over 1segments, 167169, 172
173
1 forms over oriented regions, 285,
292 (Ex. 4)
2forms over 2segments, 169172,
174 (Ex. 8), 297 (Ex. 21)
Interior angle, 374
Intrinsic distance, 264265, 269(Ex. 3),
351
Intrinsic geometry, 271, 304
Inverse function, 2
Inverse function theorem, 39, 161162
Isometric imbedding, 367369
Isometric immersion, 367369
Isometric invariant, 271, 304
Isometric surfaces, 265
Isometry of Euclidean space, 98111
decomposition theorem, 101
derivative map, 104
determined by frames, 105
Isometry group, 365
Isometry of surfaces, 263266, 270275,
306
Euclidean isometries and, 297299
isometric immersions and, 367
Isothermal coordinates, 279 (Ex. 2)
J (rotation operator), 309(Ex. 5)
Jacobian, 288, 294(Ex. 8), 296(Ex. 17)
Jacobian matrix, 37
Jacobi equation, 355357, 361 (Ex. 6)
K
Kronecker delta, 23, 45
L
2, m, n, 211213
Lagrange identity, 20® (Ex. 6)
Law of cosines, 370 (Ex. 10)
Leibnizian property, 13
Length
of a curve segment, see Arc length
of a vector, 44
Levi Civita, 322
Liebmann's theorem, 262
Line of curvature, see Principal curve
Liouville parametrization, 339 (Ex. 18)
Local isometry, 265270, 362365
of constant curvature surfaces, 364
criteria for, 266, 269(Ex. 1), 270(Ex. 10)
determined by frames, 363
Local minimization of arc length, 352358
M
Manifold, 184187
Mapping
of Euclidean spaces, 3241
of surfaces, 158166
Mean curvature, 203, 205208, 212, 217,
252
Mercator projection, 271 (Ex. 13)
Metric tensor, 305
Minimal surface, 207, 275(Ex. 5)
examples, 238
flat, 263(Ex. 1)
Gauss mapping, 294(Ex. 6), 296(Ex. 20)
of revolution, 236238
ruled, 233 (Ex. 22)
Minimization of arc length, 340
Mobius band, 178, 180(Ex. 7)
Monge patch, 127, 219
Monkey saddle, 132(Ex. 6), 205, 231 (Ex.
9)
Gaussian curvature, 220(Ex. 5)
N
Natural coordinate functions, 4
Natural frame field, 9
Neighborhood, 43, 125
normal, 341
Norm, 43
Normal curvature, 195202, 221 (Ex. 14)
sign of, 198
Normal section, 197
Normal vector field, 147
npolygon, 388389
O
1 segment, 167
Onetoone, 2
INDEX
409
Onto, 2
Open interval, 15
Open set, 5, 43, 151152
Orientable surface, 177178, 185(Ex. 1)
Orientation
determined by a patch, 374
of a frame, 107
of a patch, 285
of a paving, 285
of a surface, 195, 284
of tangent frame fields, 291
Orientationpreserving (reversing) isom
etry, 109, 296
Orientationpreserving (reversing) re
parametrization, 52
Oriented angle, 291
Orthogonal coordinates, 276280
Gaussian curvature formula, 278
Orthogonal matrix, 46
Orthogonal transformation, 100
Orthogonal vectors, 44
Orthonormal expansion, 4546, 83
Orthonormal frame, see Frame
Osculating circle, 64(Ex. 6)
Osculating plane, 61
Parallel curves, 116
Parallel postulate, 336
Parallel surfaces, 209210
Parallel translation, 323
Parallel vector field, 5455, 322
Parallel vectors, 6
Parameter curves, 133
Parametrization of a curve, 21
Parametrization in a surface, 135
criteria for regularity, 136, 140(Ex. 2)
decomposable into patches, 165(Ex. 9)
Partial velocities, 134, 136, 146
Patch, 124137
abstract, 182, 184
geometric computations, 210216
Monge, 127
orthogonal, 220 (Ex. 9), 276280
principal, 220(Ex. 9), 280(Ex. 4)
proper, 124, 152 (Ex. 14)
Patchlike 2segment, 281
Paving, 283, 377
Planar point, 205
Plane in E 3 , 60, 131 (Ex. 2), 228, 256
identified with E 2 , 126
Poincare 1 halfplane, 309(Ex. 2)
Gaussian curvature, 317 (Ex. 1)
geodesies, 337 (Ex. 6)
isometric to hyperbolic plane, 371 (Ex.
15)
polar circles, 351 (Ex. 2)
Point of application, 6
Pointwise principle, 8
Polar circle, 346. 359360
Polar disc, 361 (Ex. 2)
Polygonal region, 386
Pregeodesic, 330
Principal curvatures, 199207
as characteristic values, 200
formula for, 206
Principal curve, 223225, 230232, 263
(Ex. 4)
Principal direction, 199
Principal frame field, 254
Principal normal, 57, 66, 69
Principal vectors, 199, 220(Ex. 10), 223
(Ex. 22)
as characteristic vectors, 200
Projective plane, 182183
geodesies, 337 (Ex. 8)
geometric structure, 317 (Ex. 6)
homogeneity, 371 (Ex. 11)
natural mapping (projection), 183
topological properties, 186(Ex. 2)
Pseudosphere, see Bugle surface
Pullback, 163
Quadratic approximation, 202204
Quadric surface, 142, 294 (Ex. 10)
R
Rectangular decomposition, 377
Reflection, 109
Regular curve, 20
Regular mapping, 38, 161
Reparametrization, 18
monotone, 56 (Ex. 10)
orientation preserving (reversing), 52
unit speed, 51
Riemann, 304, 336
Riemannian geometry, 308, 389390
Riemannian manifold, 308
Rigid motion, see Isometry of Euclidean
space
Rigidity, 301 (Ex. 1)
410
INDEX
Rotation, 111 (Ex. 4)
Ruled surface, 140143, 227, 231233
noncylindrical, 232(Ex. 14)
total Gaussian curvature, 294 (Ex. 9)
Ruling, 140
Saddle surface, 192
doubly ruled, 227
Euclidean symmetries, 303 (Ex. 9)
patch computations, 214216
principal vectors, 221 (Ex. 11)
Scalar multiplication, 3, 89
Scale factor, 268
Scherk's surface, 222(Ex. 21)
patch in, 303 (Ex. 11)
Gauss mapping, 296 (Ex. 20)
Schwarz inequality, 44
Serret, 81
Shape operator, 190194
characteristic polynomial, 208 (Ex. 4)
co variant derivatives and, 324 (Ex. 3)
as derivative of Gauss mapping, 289
frame fields, in terms of, 248
Gaussian and mean curvature, and, 203
of an immersed surface, 368
normal curvature and, 196
preserved by Euclidean isometries,
297298
principal curvatures and vectors, and,
200
proof of symmetry, 212213, 251 (Ex. 6)
Shortest curve segment, 340
Sign of an isometry, 108
Simple region, 294295
Simply connected surface, 176, 363
Slant of a geodesic, 331, 338
Smooth overlap, 145, 182
Speed, 51
Sphere, 128
conjugate points, 354, 357358
Euclidean symmetries, 302(Ex. 8)
framehomogeneity, 370 (Ex. 6)
Gaussian curvature, 207, 219(Ex. 1),
253, 360, 361 (Ex. 2)
geodesies, 228229, 346347
geographical patch, 134135, 277278
geometric characterizations, 258, 259,
262
geometric structures, 383
with handles, 379
holonomy, 323324, 337 (Ex. 4)
local isometries, 363365
rigidity, 301 (Ex. 1)
shape operator, 191192
topological properties, 176178, 378
Spherical curve, 63, 65 (Ex. 10)
Spherical frame field, 83
adapted to sphere, 248249, 277278
dual and connection forms, 9495
Spherical image
of a curve, 71, 75 (Ex. 11)
of a surface, see Gauss mapping
Standard geometric surface, 363366
Stereographic plane, 314
Stereographic projection, 160, 162
as conformal mapping, 271 (Ex. 14)
Stereographic sphere, 314, 337 (Ex. 5)
Stokes' theorem, 170172, 387 (Ex. 7)
Straight line, 15, 18, 55, 229(Ex. 1)
lengthminimizing properties, 56(Ex.
1)
Striction curve, 232 (Ex. 14)
Structural equations
on E 3 , 9295
on a surface, 249, 252, 292, 311312
Subset, 1
Support function, 218, 222, 256
Surface
abstract, 182184
in E 3 , 125, 306, 367
implicit definition, 127128
geometric, 305
immersed, 368
Surface of revolution, 129130, 234244
area, 292 (Ex. 2)
augmented, 133 (Ex. 12)
of constant curvature, 239241, 244
(Ex. 9), 294 (Ex. 7)
diffeomorphism types, 187(Ex. 8)
Gaussian curvature, 235, 238, 242,
243(Ex. 5)
geodesies, 338(Ex. 13)
local characterization, 270(Ex. 12)
meridians and parallels, 130
natural frame field, 279
parametrization
canonical, 238
special, 143 (Ex. 13)
usual, 138139
principal curvatures, 235
principal curves, 225, 235
INDEX
411
topological properties, 182(Ex. 14)
total curvature, 293 (Ex. 5)
Symmetry equation, 249
Tangent bundle, 185
Tangent line, 22(Ex. 9)
Tangent plane, 146, 150(Ex. 9)
Tangent space, 7
Tangent surface, 231 (Ex. 11)
isometries of, 270(Ex. 5), 301 (Ex. 2)
Tangent vector
to E 3 , 6, 14
to a surface, 146, 183184
Theorema egregium, 273275
Topological properties, 176182, 380n
Toroidal frame field, 84(#x. 4), 95 (Ex. 2),
250251
Torsion, 58, 66
formula, 69
sign, 114115
Torus of revolution, 139
EulerPoincar£ characteristic, 379, 387
(Ex. 9)
Gaussian curvature, 204205, 235236
Gauss mapping, 194(Ex. 5), 290291
patch computations, 235236
total Gaussian curvature, 287, 291
usual parametrization, 139
Total curvature of a curve, 7Q(Ex. 16)
Total Gaussian curvature, 286291, 380
385
EulerPoincare 1 characteristic, and, 380
Gauss mapping, and, 290
holonomy and, 325 (.Re. 5)
of a patch, 325(Ex. 5)
Total geodesic curvature, 372375, 386
(Ex. 4),389CEr. 15)
Transferred frame field, 272273
Translation, 98100, 109
Triangle, 383385
Triangle inequality, 347
Triple scalar product, 4849, 108
Tube, 221 (Ex. 16)
2segment, 169
U
Umbilic point, 200, 221 (Ex. 13), see also
Allumbilic surface
Unit normal function, 211, 368
Unit normal vector field, l80(Ex. 5), 190
Unit points, 34
Unit speed curve, 51
Unit sphere, 125
Unit tangent, 56, 66, 69
Unit vector, 44
Vector, see Tangent vector
Vector analysis, 31 (Ex. 8)
Vector field
on an abstract surface, 183
on a curve, 5254, 320
on Euclidean space, 8
on a surface in E 3 , 147149, 151 (Ex. 12)
normal, 147, 149
tangent, 147, 149
Vector part, 6
Velocity, 1718, 183184
Vertices, 373374
W
Wedge product, 2728, 153
Winding number, 174(Ex. 5)
Open
University
Course
Units 06
M334
Differential
Geometry
Q C tf 334 PART * THE OPEN UNIVERSITY
; Mathematics: A Third Level Course
516. j
Jr
DIP
PARTO GUIDE TO THE COURSE
9
DIFFERENTIAL GEOMETRY
9
THE OPEN UNIVERSITY
Mathematics: A Third Level Course
DIFFERENTIAL GEOMETRY PART
O.U. COURSE UNITS COLLECTION
FOR REFEu£.«CE ONLY
■ — GUIDE TO THE COURSE
Prepared by the Course Team
THE OPEN UNIVERSITY PRESS
Course Team
M334
Chairman:
Mr. P.E.D. Strain
Dr. R.A. Bailey
Lecturer in Mathematics
Course Assistant in Mathematics
With assistance from:
Dr.J.M. Aldous
Mr. G.J. Burt
Dr. P.M. Clark
Mr. P.B. Cox
Dr. F.C. Holroyd
Mr. T.C. Lister
Mr. R.J. Margolis
Dr. C.A. Rowley
Mr. M.G. Simpson
Senior Lecturer in Mathematics
Lecturer in Educational Technology
Lecturer in Physics
Student Computing Service
Lecturer in Mathematics
Staff Tutor in Mathematics
Staff Tutor in Mathematics
Course Assistant in Mathematics
Course Assistant in Mathematics
Consultants:
Prof. S. Robertson
Prof. T. Willmore
Professor of Pure Mathematics,
University of Southampton
Professor of Pure Mathematics,
University of Durham
.. ;•; b°\
3
;LASSN ^I6^^ "^^
/'
LOAN CAT Kf\^ ""^r
The Open University Press, Walton Hall, Milton Keynes
First published 1975
Copyright © 1975 The Open University
All rights reserved. No part of this work may be reproduced in any form, by mimeograph or any other means
without permission in writing from the publishers.
Produced in Great Britain by
The Open University Press
ISBN 335 05700 4
This text forms part of the correspondence element of an Open University Third Level Course. The complete
list of parts in the course is given at the end of this text.
For general availability of supporting material referred to in this text, please write to the Director of
Marketing, The Open University, P.O. Box 81, Milton Keynes, MK7 6AT.
Further information on Open University courses may be obtained from The Admissions Office, The Open
University, P.O. Box 48, Milton Keynes, MK7 6AB.
1.1
M334
CONTENTS Page
A Brief History of Differential Geometry 5
Bibliography and Conventions 6
Prerequisites 7
Prerequisite Sections 7
Detailed Prerequisites 8
Handbooks 10
The Structure of the Course 11
1. Breakdown into Blocks 11
2. Structure of Collection of Blocks 12
3. Detailed Structure of Blocks and Interconnections 12
A possible 16week Breakdown of the Course 16
Working Instructions 1 7
Further Reading 2 1
Errata 22
M334
A BRIEF HISTORY OF DIFFERENTIAL GEOMETRY
This is a very brief history of those aspects of differential geometry that are dealt
with in this course. It is here to give some historical perspective in a course which
developes the modern approach to old questions. Differential geometry is a vast
subject and most of the great mathematicians of the last three centuries have made
some contribution to it.
Some of the basic problems of differential geometry were already being studied
before the introduction of the calculus. For instance, in connection with his study
of light, Huygens (162995) studied the curvature of certain plane curves. However,
the calculus enabled Newton (16471727) to make a more detailed study of such
curves and Euler (170783) was able to study curves on more general surfaces. The
study of surfaces themselves was opened up by the use of patches introduced in a
restricted way by Monge (17461818) and used by Gauss (17771855) to describe
the curvature of a surface in an intrinsic way. This enabled one to answer such
questions as whether it was possible to make a plane map of the globe that retained
all the local properties of its geometry.
It was Riemann (182666) who saw how the theory of surfaces could be generalized.
The study of Riemann surfaces showed finally that Euclidean geometry was not the
only possibility. There is now a vast literature on such geometries.
Meanwhile, the theory of curves took a great step forward with the discovery, by
Frenet (18161900) and Serret (181985), of formulas describing the curvature and
torsion of curves. Their "method of moving frames" was adapted by Darboux
(18421907) to describe surfaces. This method was brought to full generality by
Cartan (18691951), who brought back geometric insight into the profusion of
equations thrown up by Riemann and his followers.
A readable introduction to some of the geometric ideas dealt with in this course can
be found in:
D. Hilbert and S. CohnVossent: Geometry and the Imagination (Chelsea, 1952).
This section on differential geometry contains pictures of several of the surfaces we
shall be studying.
b M334
BIBLIOGRAPHY AND CONVENTIONS
Bibliography
The set book for this course is Barrett O'Neill: Elementary Differential Geometry
(Academic Press, 1966). It is essential to have this book: the course is based on it
and will not make sense without it.
The set books for M201, M231 and MST 282 are referred to occasionally; they are
useful but not essential. They are:
D.L. Kreider, R.G. Kuller, D.R. Ostberg and F.W. Perkins: An Introduction to
Linear Analysis (AddisonWesley, 1966).
E.D. Nering: Linear Algebra and Matrix Theory (John Wiley, 1970).
M. Spivak: Calculus, paperback edition (W.A. Benjamin/ AddisionWesley, 1973).
R.C. Smith and P. Smith: Mechanics, SI Edition (John Wiley, 1972).
Conventions
Unreferenced pages and sections denote the set book. Otherwise
O'Neill denotes the set book;
Text denotes the correspondence text;
KKOP denotes An Introduction to Linear Analysis by D.L. Kreider,
R.G. Kuller, D.R. Ostberg and F.W. Perkins;
Nering denotes Linear Algebra and Matrix Theory by E.D. Nering;
Spivak denotes Calculus by M. Spivak;
Smith denotes Mechanics by R.C. Smith and P. Smith.
Reference to Open University Courses in Mathematics take the form:
Unit Ml 00 22, Linear A Igebra I
Unit MST 281 10, Taylor Approximation
Unit M2 01 16, Euclidean Spaces I: Inner Products
Unit M231 2, Functions and Graphs
Unit MST 282 1, Some Basic Tools.
References to Definitions, Theorems, Lemmas etc. in the set book, O'Neill, are given
as in the following example :
Lemma V.4.2 stands for Lemma 4.2 of Chapter V.
Sometimes if O'Neill is referring to something in the same chapter he omits the
roman numeral for the Chapter.
M334
PREREQUISITES
The courses that are prerequisites for this course are M201 Linear Mathematics,
M231 Analysis, and MST 282 Mechanics and Applied Calculus.
However, this course has been written with very few explicit references to any of
these courses. In this section we list those units and sections of the prerequisite
courses that we have built on, and this is followed by a brief description of their
content. This should enable you to work out whether you are familiar with those
ideas and results that we shall be assuming. The bracketed sections below are marked
as optional to M231 and so you may not have read them yet. If you want to, do so
now, noting only the ideas and results. The details of the definitions and proofs will
not be needed and our notation will not be similar.
Prerequisite Sections
M201 Linear Mathematics
Unit 1 Vector Spaces
Unit 2 Linear Transformations
Unit 3 Hermite Normal Form, Section 3.3
Unit 5 Determinants and Eigenvalues
Unit 10 Jordan Normal Form, Section 10.2.1
Unit 12 Linear Functionals and Duality, Section 12.1
Unit 14 Bilinear and Quadratic Forms, Sections 14.1 and 14.2.3
Unit 15 Affine Geometry and Convex Cones, Sections 15.1.1 and 15.1.2
Unit 16 Euclidean Spaces I, Inner Proudcts
Unit 23 The Wave Equation, Section 23.1.1
Unit 24 Orthogonal and Symmetric Transformations, Sections 24.1, 24.3.1 and
24.3.2
M231 Analysis
Unit 2 Functions and Graphs (plus Section 2.7)
Unit 5 Differentiation (plus Section 5.9)
(Unit 7 Applications of the Derivative, Section 7.8)
Unit 10 Some Important Functions
MST 282 Mechanics and Applied Calculus
Unit 1 Some Basic Tools, Sections 1.1. to 1.3
Unit 2 Kinematics, Sections 2.1.1 and 2.2
Unit 7 Work and Energy I, Section 7.1.2
8 M334
Detailed Prerequisites
M201 Linear Mathematics
This is the major prerequisite for our course. Though we do not often quote results
from it, we freely use all the basic ideas and conventions that it introduced.
Unit 1 Vector Spaces This unit introduces the vector space of arrows, R 2 and
R 3 , which are the basis for our course. You should also know about linear
dependence, bases and coordinates, though we shall deal only with concrete
examples in R 3 . Ideas such as dimension and subspaces are implicit in our course but
you will not need any explicit results. In fact, the ideas of this unit are basic to all
linear mathematics.
Unit 2 Linear Transformations This unit again introduces ideas basic to our
course. You will need to know about linear transformations , the convention for
representing them by matrices, isomorphisms and rank, though we shall be in
terested in 2 and 3dimensional examples only. This is another unit with which to
be entirely familiar.
Unit 3 Hermite Normal Form We do not use the technique of Hermite normal
forms in our course. However, we do talk about the rank of 2 X 2 and 3X3
matrices, which can be read off the Hermite normal form obtained by the elemen
tary operations in Section 3.3.
Unit 5 Determinants and Eigenvalues Determinants, eigenvalues and eigen
vectors play a small but important part in our course. The cross product of two
vectors will be defined in terms of 3 X 3 determinants and you should be familiar
with the formula for expanding such a determinant dealt with in KKOP, pages
682  687, where the effect of column operations is also described. In Section 5.1.5 it
is shown that row operations behave similarly and it is these that we shall use. You
will need to know only what eigenvectors and eigenvalues are and not in general how
to calculate them. We shall be using eigenvector bases as in Section 5.2.4.
Unit 10 Jordan Normal Form In Section 10.2.1 it is shown that a linear trans
formation is represented by a diagonal matrix with respect to an eigenvector basis.
Unit 12 Linear Functional and Duality It will be useful to know the definition
of linear functionals, the dual space and a dual basis. However, do not worry about
any technical results. Since we shall be dealing with R 2 and R 3 , in which there is a
simple inner (scalar, dot) product, we shall be able to describe dual bases etc. in a
concrete way.
Unit 14 Bilinear and Quadratic Forms Again you will find it useful to have met
bilinear forms and symmetric bilinear forms. It will give you some more insight into
the course, although the only explicit result we shall use is the quadratic Taylor
approximation.
Unit 15 A f fine Geometry and Convex Cones Though this unit is not in the spirit
of our course, Sections 15.1.1 and 15.1.2 give various ways of describing affine lines
and planes which may prove useful to you.
M334
PREREQUISITES
The courses that are prerequisites for this course are M201 Linear Mathematics,
M231 Analysis, and MST 282 Mechanics and Applied Calculus.
However, this course has been written with very few explicit references to any of
these courses. In this section we list those units and sections of the prerequisite
courses that we have built on, and this is followed by a brief description of their
content. This should enable you to work out whether you are familiar with those
ideas and results that we shall be assuming. The bracketed sections below are marked
as optional to M231 and so you may not have read them yet. If you want to, do so
now, noting only the ideas and results. The details of the definitions and proofs will
not be needed and our notation will not be similar.
Prerequisite Sections
M201 Linear Mathematics
Unit 1 Vector Spaces
Unit 2 Linear Transformations
Unit 3 Hermite Normal Form, Section 3.3
Unit 5 Determinants and Eigenvalues
Unit 10 Jordan Normal Form, Section 10.2.1
Unit 12 Linear Functionals and Duality, Section 12.1
Unit 14 Bilinear and Quadratic Forms, Sections 14.1 and 14.2.3
Unit 15 Affine Geometry and Convex Cones, Sections 15.1.1 and 15.1.2
Unit 16 Euclidean Spaces I, Inner Proudcts
Unit 23 The Wave Equation, Section 23.1.1
Unit 24 Orthogonal and Symmetric Transformations, Sections 24.1, 24.3.1 and
24.3.2
M231 Analysis
Unit 2 Functions and Graphs (plus Section 2.7)
Unit 5 Differentiation (plus Section 5.9)
(Unit 7 Applications of the Derivative, Section 7.8)
Unit 10 Some Important Functions
MST 282 Mechanics and Applied Calculus
Unit 1 Some Basic Tools, Sections 1.1. to 1.3
Unit 2 Kinematics, Sections 2.1.1 and 2.2
Unit 7 Work and Energy I, Section 7.1.2
8 M334
Detailed Prerequisites
M201 Linear Mathematics
This is the major prerequisite for our course. Though we do not often quote results
from it, we freely use all the basic ideas and conventions that it introduced.
Unit 1 Vector Spaces This unit introduces the vector space of arrows, R 2 and
R 3 , which are the basis for our course. You should also know about linear
dependence, bases and coordinates, though we shall deal only with concrete
examples in R 3 . Ideas such as dimension and subspaces are implicit in our course but
you will not need any explicit results. In fact, the ideas of this unit are basic to all
linear mathematics.
Unit 2 Linear Transformations This unit again introduces ideas basic to our
course. You will need to know about linear transformations, the convention for
representing them by matrices, isomorphisms and rank, though we shall be in
terested in 2 and 3dimensional examples only. This is another unit with which to
be entirely familiar.
Unit 3 Hermite Normal Form We do not use the technique of Hermite normal
forms in our course. However, we do talk about the rank of 2 X 2 and 3X3
matrices, which can be read off the Hermite normal form obtained by the elemen
tary operations in Section 3.3.
Unit 5 Determinants and Eigenvalues Determinants, eigenvalues and eigen
vectors play a small but important part in our course. The cross product of two
vectors will be defined in terms of 3 X 3 determinants and you should be familiar
with the formula for expanding such a determinant dealt with in KKOP, pages
682 _ 687, where the effect of column operations is also described. In Section 5.1.5 it
is shown that row operations behave similarly and it is these that we shall use. You
will need to know only what eigenvectors and eigenvalues are and not in general how
to calculate them. We shall be using eigenvector bases as in Section 5.2.4.
Unit 10 Jordan Normal Form In Section 10.2.1 it is shown that a linear trans
formation is represented by a diagonal matrix with respect to an eigenvector basis.
Unit 12 Linear Functional and Duality It will be useful to know the definition
of linear functionals, the dual space and a dual basis. However, do not worry about
any technical results. Since we shall be dealing with R 2 and R 3 , in which there is a
simple inner (scalar, dot) product, we shall be able to describe dual bases etc. in a
concrete way.
Unit 14 Bilinear and Quadratic Forms Again you will find it useful to have met
bilinear forms and symmetric bilinear forms. It will give you some more insight into
the course, although the only explicit result we shall use is the quadratic Taylor
approximation.
Unit 15 A f fine Geometry and Convex Cones Though this unit is not in the spirit
of our course, Sections 15.1.1 and 15.1.2 give various ways of describing affine lines
and planes which may prove useful to you.
M334 9
Unit 16 Euclidean Spaces I: Inner Products Sections 16.1 to 16.3 are very
important to our course. They introduce the inner (scalar, dot) product of vectors,
which we use extensively. You will need to be able to use it to define length,
distance, angle and orthonormal bases. We shall frequently use orthonormal bases,
orthonormal expansions and the simple form of the inner product in terms of the
coordinates with respect to such a basis, all of which are dealt with in Section
16.4.3. Our course goes over all the details again.
Unit 23 The Wave Equation All you need from this unit is the definition of
partial differention in Section 23.1.1, though even this will be redefined.
Unit 24 Orthogonal and Symmetric Transformations This unit contains many
results that are needed in this course. Orthogonal transformations are linear space
'morphisms' and the orthogonal matrices representing them are characterized in
several important ways, including the result that their inverse is equal to their
transpose. We shall also use the result that symmetric transformations can be repre
sented by diagonal matrices with respect to suitable eigenvector bases.
M231 Analysis
Unit 2 Functions and Graphs This unit covers all the basic ideas associated with
graphs and functions of one variable with which you need to be familiar.
The appendix, Section 2.7, is very straightforward and you ought to have a look at
it. It introduces ideas that will be developed in this course.
Unit 5 Differentiation This unit introduces the basic definitions of differentia
tion and curves. These are ideas which will be developed further in this course.
However, our approach will be less theoretical. We shall be interested in using
techniques rather than worrying about such problems as differentiability.
We shall also be using partial and directional derivatives extensively. These are intro
duced in an appendix, Section 5.9. If you have looked at this before you might like
to look at it again, otherwise don't worry.
Unit 7 Applications of the Derivative The only relevant part of this unit is the
appendix, Section 7.8, on implicit functions. This is straightforward and will help
you with Chapter IV of this course. Try to look through it before then but do not
bother with any proofs or the sections on contunuity and differentiability.
Unit 10 Some Important Functions This unit introduces all the standard func
tions that you will need in this course. You will need to know about polynomials,
trigonometric functions, the exponential and logarthmic functions, hyperbolic
functions and inverse trigonometric and hyperbolic functions. Useful relations in
volving these and their derivatives can be found in the summary, Section 10.10.
MST 282 Mechanics and Applied Calculus
Unit 1 Some Basic Tools Sections 1.1 to 1.3 give a useful introduction to
vectors and applied calculus, with which you should be familiar. We shall use the
scalar, vector and triple scalar products of vectors and some simple forms of the
chain rule for differentiation.
10 M334
Unit 2 Kinematics In Section 2.1.1 differentiation of a vector is introduced and
in Section 2.2 parametric equations of curves are discussed. These are ideas that will
be developed further and so you should be familiar with them.
Unit 7 Work and Energy I Section 7.1.2 contains some very important ideas
that we shall develop further and so you should be familiar with them. It deals with
scalar fields, level curves and surfaces, directional derivatives and the gradient. It also
contains a simple form of the chain rule for differentiation, for which it gives a
proof in Appendix 1.
Handbooks
Each of the prerequisite courses has a handbook which contains some of the defini
tions to which we refer. However, the most useful parts of the handbooks are those
giving lists of the derivatives of standard functions. These are
M231 Handbook: Section 5, Familiar Functions, pages 4548
MST 282 Handbook: Section 4, Table of Derivatives, page 27.
M334
11
THE STRUCTURE OF THE COURSE
This course is based on B. O'Neill's book Elementary Differential Geometry pub
lished by Academic Press. Again we must stress that it is essential to have this book
since the course is based almost entirely on it. The course consists of the first six
chapters of the book, apart from a few odd sections, and, with one minor exception,
you should read the book in the order it is written. We expect you to read it section
by section and we have provided a commentary on each relevant section. The
commentaries have been bound together chapter by chapter.
There is another way of viewing the structure of the course which might prove
useful. All those sections dealing with a particular theme can be thought of as
forming a block. We shall show which sets of sections form blocks and also how
these blocks are interrelated, that is, which blocks it is essential to have tackled, at
least in part, before going onto another block. If you ever become stuck with the
sections in one block this will enable you to see if there is any alternative block that
you can tackle first.
Sometimes a particular section does not depend on all those sections that have
preceded it in that block or on all the sections of preceding blocks. We have included
this information in the form of a detailed structure diagram for each block giving all
prerequisite blocks and sections.
Each introduction to a section again lists the most important prerequisite sections
but we hope that if you refer to the structure diagrams below regularly the overall
structure of the course will become clearer.
1. Breakdown into Blocks
Here is the content of the blocks and the number of units to which they are
equivalent, assuming that this is a standard x h credit course with fourteen teaching
units and two gaps.
B.
D.
E.
H.
Calculus and Linear Algebra in E 3
1.1 to 1.4, 1.7, II. 1, II.2, II.5, II.6
Basic Exterior Calculus in E 3
1.5 and 1.6
Connection Forms for Frame Fields in E 3
II. 7 and II. 8
Frenet Formulas for Curves in E 3
II. 3 and II.4
Geometry of E 3
III.l to III. 3
Geometry of Curves in E 3
III.4 and III.5
Calculus on Surfaces in E 3
1V.1 to IV.3, IV.5(a)
Exterior Calculus on Surfaces in E 3
IV.4, IV.5(b), IV.7(a)
3 units
x h unit
x h unit
1 unit
x h unit
x h unit
\ x /z units
1 unit
12 M334
I. Shape Operator of a Surface in E 3
V.ltoV.4 2 units
J. Applications of the Shape Operator
V.5 and V.6 1 unit
K. Structure of Surfaces in E 3
VI. 1, VI.2,VL6 1 unit
L. Global results on Surfaces in E 3
VI.3, IV.7(b) Vi unit
M. Geometry of Surfaces in E 3
VI.4, VI.5, VI.8 1 unit
2. Structure of Collection of Blocks
The blocks are related as below, a block is preceded by all those blocks above it to
which it is connected.
So, for instance, even if you find block C too difficult, and on first reading it is quite
difficult, there is no reason why you should not press on with D, E, F, G, H, I, J as
long as you have looked at C again before you come to K that depends on it.
Each block and all those that precede it tell an interesting story. For instance the
line terminating at J consists of the blocks A, D, G, I, J and describes the shape of
surfaces of revolution and interesting curves on surfaces.
3. Detailed Structure of Blocks and Interconnections
In the following diagrams sections in the same block are grouped together. Arrows
indicate how the sections are interconnected. We have shown only how a section
depends on those directly preceding it. They in turn may depend on others.
M334
13
1.1
i
1.2
1
1.3
1
1.4
1
1.7
B
II.2
II.6
II.5
A
1.3
1.5
1
1.6
B
D
A
II.2
II.3
1
II.4
D
II. 1
A
1.7
II]
1 _
'
'
.1
11
I
1.2
* 111. J
D
II.3
II.4
E
III.3
* III.4
1
^ III.5
14
M334
H
G
IV.3
D
II.3
'
'
^ '
V
.1
■*• V
.z
I
* v .5
w V .4
M334
15
M
G
IV.5
l
H
IV.5
K
VI. 1
F
III.5
\
/
J
V.6
i

'
'
Vl4
*■ \
►
\n q
i
M
III.2
E
16
M334
A POSSIBLE 16WEEK BREAKDOWN OF THE COURSE
In case you wish to break down the course into 16 weeks of work, the following is
a suggested way of doing so.
Week
Sections of O'Neill
Blocks
TMA
1
1.1,1.2,1.3,1.4
A
2
1.5,1.6,1.7
A, B
01
3
11.1,11.2
A
4
11.3,11.4
D
5
11.5,11.6,11.7,11.8
A,C
6
02
7
III.l, IIL2, III.3, III.4, III.5
E,F
8
IV.1,IV.2
G
9
IV.3,IV.4
G,H
03
10
IV.5, IV.7(a)
G,H
11
V.1,V.2,V.3
I
12
V.4
I
13
V.5, V.6
J
04
14
VI.l,VI.2,VI.3,IV.7(b)
K, L
15
VI.4,VI.5,VI.6
K,M
16
VI. 8
M
M334 17
WORKING INSTRUCTIONS
This section deals with instructions on how to tackle written material of this course.
Of course, we realize that by now that you will probably have worked out which
way it is best for you to tackle an Open University text and you may not wish to
change successful habits. However, while we were writing our commentaries we had
to keep in mind a certain approach to reading them and this was responsible for
them being structured in the way they are.
Components
This course has two components. The first is the Set Book, B. O'Neill: Elementary
Differential Geometry, and the second is the Course Notes. The set book is divided
into Chapters which correspond to Parts of the course notes and both of these are
divided into Sections indexed in the same way.
Reading a Section
Each section is divided up in the following way.
Introduction
Each section begins with a short introduction. The content of this introduction
depends on just how much of an introduction there is in the set book. The intro
duction tells you the prerequisite sections and possibly some essential results that
will be used. It goes on to describe very briefly the content of the section, in terms
of those that have gone before, and how this section fits into the overall structure of
the course. It will also tell you if it is possible to put this section off till some later
time.
Reading Passage
This will usually be an entire section of the set book and it should be read after the
introduction. We expect you to read it more than once.
First Reading At the first reading you should try to browse through all the
reading passage and obtain an overall view. Look carefully at each definition and try
to think of examples. This is a very geometric course so you should try to draw lots
of pictures for yourself. Look carefully at any Theorems or Lemmas, making sure
you understand all the terms on both sides of each equation, but at this reading do
not worry too much about detailed proofs. You should also try and follow the steps
in any straightforward worked examples. There are several manipulative techniques
that are demonstrated in this way and you will need to master them.
18 M334
Comments
After completing the reading passage you should next look at the comments we have
provided. Though they are related to the text you are not expected to jump back
wards and forwards from the reading page to them. In fact some of the comments
are quite general and relate to an idea that is developed throughout the section
rather than some specific result. You should read these comments in the same spirit
as the reading passage.
Additional Text
The Additional Text should be read directly after the Reading Passage and the
Comments and approached in exactly the same way. The Additional Text consists of
definitions and results that we feel should be discussed at this point and it should be
treated as an integral part of the section.
Optional Text
Some sections will contain an Optional Text. These consist either of passages of the
text book or of the course notes and develop certain straightforward topics that we
consider interesting but peripheral to the main structure of the course. We have
included these Optional Texts for those of you who are finding the course enticing
and have time'to spare; others may omit them without losing anything.
Text Exercises
At the end of the Comments, Additional or Optional Texts you may find some Text
Exercises. These are included at this point because they form an integral part of the
text. They may deal with some result that has already been referred to or with some
result that will be of crucial importance later on.
Second Reading After having worked once through the Reading Passage, Com
ments and Additional Text you should work through them again but this time in
more detail. Once you have grasped the overall structure of the section you need to
fill in the finer details. Look up references needed to complete a proof and work
through all proofs and examples filling in the missing steps.
Supplementary Comments
It is not essential that you read these comments. They are intended only to help you
at the second reading. They are intended to cover those tricky or routine points
which are not essential to the development of the subject but may nevertheless cause
some concern. Anything not covered by either the Comments or the Supplementary
Comments has been omitted because we do not consider it of any importance.
Comments Versus Supplementary Comments The main difference between
these two types of comments is that Comments are an integral part of the text and
should be included at each reading and during revision, while Supplementary
Comments need be looked at only once to clear up tricky points and are not
essential to revision.
M334 17
WORKING INSTRUCTIONS
This section deals with instructions on how to tackle written material of this course.
Of course, we realize that by now that you will probably have worked out which
way it is best for you to tackle an Open University text and you may not wish to
change successful habits. However, while we were writing our commentaries we had
to keep in mind a certain approach to reading them and this was responsible for
them being structured in the way they are.
Components
This course has two components. The first is the Set Book, B. O'Neill: Elementary
Differential Geometry, and the second is the Course Notes. The set book is divided
into Chapters which correspond to Parts of the course notes and both of these are
divided into Sections indexed in the same way.
Reading a Section
Each section is divided up in the following way.
Introduction
Each section begins with a short introduction. The content of this introduction
depends on just how much of an introduction there is in the set book. The intro
duction tells you the prerequisite sections and possibly some essential results that
will be used. It goes on to describe very briefly the content of the section, in terms
of those that have gone before, and how this section fits into the overall structure of
the course. It will also tell you if it is possible to put this section off till some later
time.
Reading Passage
This will usually be an entire section of the set book and it should be read after the
introduction. We expect you to read it more than once.
First Reading At the first reading you should try to browse through all the
reading passage and obtain an overall view. Look carefully at each definition and try
to think of examples. This is a very geometric course so you should try to draw lots
of pictures for yourself. Look carefully at any Theorems or Lemmas, making sure
you understand all the terms on both sides of each equation, but at this reading do
not worry too much about detailed proofs. You should also try and follow the steps
in any straightforward worked examples. There are several manipulative techniques
that are demonstrated in this way and you will need to master them.
1Q M334
lo
Comments
After completing the reading passage you should next look at the comments we have
provided. Though they are related to the text you are not expected to jump back
wards and forwards from the reading page to them. In fact some of the comments
are quite general and relate to an idea that is developed throughout the section
rather than some specific result. You should read these comments in the same spirit
as the reading passage.
Additional Text
The Additional Text should be read directly after the Reading Passage and the
Comments and approached in exactly the same way. The Additional Text consists of
definitions and results that we feel should be discussed at this point and it should be
treated as an integral part of the section.
Optional Text
Some sections will contain an Optional Text. These consist either of passages of the
text book or of the course notes and develop certain straightforward topics that we
consider interesting but peripheral to the main structure of the course. We have
included these Optional Texts for those of you who are finding the course enticing
and have time to spare; others may omit them without losing anything.
Text Exercises
At the end of the Comments, Additional or Optional Texts you may find some Text
Exercises. These are included at this point because they form an integral part of the
text. They may deal with some result that has already been referred to or with some
result that will be of crucial importance later on.
Second Reading After having worked once through the Reading Passage, Com
ments and Additional Text you should work through them again but this time in
more detail. Once you have grasped the overall structure of the section you need to
fill in the finer details. Look up references needed to complete a proof and work
through all proofs and examples filling in the missing steps.
Supplementary Comments
It is not essential that you read these comments. They are intended only to help you
at the second reading. They are intended to cover those tricky or routine points
which are not essential to the development of the subject but may nevertheless cause
some concern. Anything not covered by either the Comments or the Supplementary
Comments has been omitted because we do not consider it of any importance.
Comments Versus Supplementary Comments The main difference between
these two types of comments is that Comments are an integral part of the text and
should be included at each reading and during revision, while Supplementary
Comments need be looked at only once to clear up tricky points and are not
essential to revision.
M334 19
Summary
After mastering as much of the Reading Passage, Comments and Additional Text as
you can, you should look through the Summary provided in the Course Notes. Of
course if you feelllike preparing your own summary and comparing it with ours you
might find it a worthwile exercise. Our summary has several subsections. First comes
Notation, then Definitions, Results, Examples and Techniques.
Notation We have included only representative examples of each notation intro
duced in this section.
Definitions and Results We have included in our list of Definitions and Results
some that O'Neill has mentioned only in passing but which we feel are important
enough to be considered on a par with all the others.
Examples We have included only those examples that are dealt with regularly
throughout the course. Important results about these examples are recorded in later
Results subsections.
Techniques These techniques could be referred to as the aims of the Sections.
For some of the techniques you need to be able to recognize some new object. For
others you need to be able to perform some straightforward manipulation, while for
others you need to be able to apply some abstract result or reproduce an argument
similar to one used in the text.
Exercises
After inspecting the Summary you should try the Exercises without looking at the
solutions. The Exercises fall into the following categories.
Text Exercises These were mentioned above.
Technique Exercises These are the most important exercises and are the type we
shall expect you to have mastered. We have related each exercise to the relevant
techniques so that you know exactly how you ought to tackle it.
Theory Exercises There will be only a few of this type of exercise. They derive
some abstract result or slightly extend those dealt with in the Section.
Optional Exercises These are based on or sometimes constitute the Optional
Text and so are only for those working through the Optional Texts.
Solutions
The solutions follow the exercises. We have tried to give fairly lengthy solutions to
the early exercises, pointing out the reasons for all the steps, but as time goes on we
omit reasoning that has been used previously.
20 M334
Further Exercises
At the end of our commentary on each Chapter there will be a selection of Further
Exercises that can be used for selfassessment or revision purposes. They will be
either Technique Exercises or Other Recommended Exercises. The techniques
needed will be those of the relevant section and each of the other Recommended
Exercises will be accompanied by a brief description of how the exercise fits into the
course. The Further Exercises will be followed by very brief Solutions.
M334
FURTHER READING
If you are interested in learning more about the subject after you have completed
the course, then there are the remaining sections for you to read.
Sections IV.6 and VI. 7 deal with integration on a surface in E 3 ; Section IV.8
introduces manifolds, and Chapter VII applies all the techniques and results of the
preceding Chapters to the study of Riemannian Geometry.
O'Neill also provides a short bibliography on page 391. For more differential geo
metry, at about the same level, you should read the book by T.J. Willmore; for a
more advanced approach you should read the book by N. J Hicks, and for applica
tions the book by H. Flanders.
There are a large number of other books now available on differential geometry and
it is also the subject of many current research papers. We hope that this course serves
you as an introduction to this literature.
22 M334
ERRATA
The following list comprises the significant mathematical errors we have found in
the text of O'Neill. It does not include typographical errors or mistakes where the
meaning is nonetheless clear. Errors in exercises and their solutions are generally
given not here but where the exercises are mentioned in the correspondence text.
Chapter I
Page 25, last line of text: The coefficient of v 3 should be {p\ + 2).
Page 31, Exercise 6: 'V in the last line should be "#".
Page 37, line 16: "Corollary 7.6" should be "Theorem 7.5".
d 2 a 2
Chapter II
Page 54, line 9: The second term of a" should be
dt 2
Page 80, Exercise 1: "VyW" should be "V V W".
Chapter III
Page 102, line 2: "q = F(q)" should be "q = F(p)".
Page 108, line 8: This should read
Page 108, lines 5, 4: The righthand side of these should read
" = det(,4*C)
= det ^4.det *C = det ^4.det C".
Page 120, line 7: The last two "F"s should be removed.
Oiapter IV
Page 136, line 5: The final term should be sinV
Page 148, Lemma 3.8: This should read
"If M: g = c is a surface in E 3 and if the gradient vector field
Vg= X (3#/d*Y) U( (considered only at points of M) is never zero, then it
is a nonvanishing normal vector field on the entire surface M."
The first three lines of the proof should be omitted.
Page 157, Exercise 5. The righthand side of the first equation should be
S'(f(p))v p [/1.
M334 23
Chapter V
Page 204, lines 9 and 2: "T p {M)" should be "Tp(M)" in each case.
Page 210, line 9: The final "V should be "x„".
Page 214, line 6: "1" should be "1/6
2»
Page 218, line 4 and diagram: "7p(iW)» should be "Tp{M)" in each case.
Page 241, line 5: The value of each curvature should be multiplied by 1.
Chapter VI
Page 252, line 18: The matrix should be transposed.
Page 297, line 6 of text: The second surface should be
M: z = * 2 y 2 .
2
Page 299, line 2: The righthand side of the equation should be
24 M334
DIFFERENTIAL GEOMETRY
I Calculus on Euclidean Space
II Frame Fields
III Euclidean Geometry
IV Calculus on a Surface
V Shape Operators
VI Geometry of Surfaces in E 3
WpW*
PART I THE OPEN UNIVERSITY
k<\c Mathematics: A Third Level Course ^^
DIP
PART I CALCULUS ON EUCLIDEAN SPACE
DIFFERENTIAL GEOMETRY
9
THE OPEN UNIVERSITY
Mathematics: A Third Level Course
D [FFERENTIAL GEOMETRY PART I
FOR REf ek:.;ce only
CALCULUS ON EUCLIDEAN SPACE
A commentary on Chapter I of O'Neill's
Elementary Differential Geometry
Prepared by the Course Team
THE OPEN UNIVERSITY PRESS
Course Team
M334I
Chairman:
Mr. P.E.D. Strain
Dr. R.A. Bailey
Lecturer in Mathematics
Course Assistant in Mathematics
With assistance from:
Dr. J.M. Aldous
Mr. G.J. Burt
Dr. P.M. Clark
Mr. P.B. Cox
Dr. F.C. Holroyd
Mr. T.C. Lister
Mr. R.J. Margolis
Dr. C.A. Rowley
Mr. M.G. Simpson
Senior Lecturer in Mathematics
Lecturer in Educational Technology
Lecturer in Physics
Student Computing Service
Lecturer in Mathematics
Staff Tutor in Mathematics
Staff Tutor in Mathematics
Course Assistant in Mathematics
Course Assistant in Mathematics
Consultants:
Prof. S. Robertson
Prof. T. Willmore
Professor of Pure Mathematics,
University of Southampton
Professor of Pure Mathematics,
University of Durham
SAK CGUc Ha
*G3
CLASS No.
The Open University Press, Walton Hall, Milton Keynes
First published 1975
Copyright © 1975 The Open University
All rights reserved. No part of this work may be reproduced in any form, by mimeograph or any other means,
without permission in writing from the publishers.
Produced in Great Britain by
The Open University Press
ISBN 335 05700 4
This text forms part of the correspondence element of an Open University Third Level Course. The complete list
of parts in the course is given at the end of this text.
For general availability of supporting material referred to in this text, please write to the Director of Marketing,
The Open University, P.O. Box 81, Milton Keynes, MK7 6AT.
Further information on Open University courses may be obtained from The Admissions Office, The Open
University, P.O. Box 48, Milton Keynes, MK7 6AB.
1.1
M334I
CONTENTS Page
Set Book 4
Bibliography 4
Conventions 4
1.1 Introduction and Euclidean Space 5
1.2 Tangent Vectors 10
1.3 Directional Derivatives 16
1.4 Curves in E 3 22
1.5 1Forms 30
1.6 Differential Forms 37
1.7 Mappings 43
Further Exercises and Solutions 54
4 M334 1
Set Book
Barrett O'Neill: Elementary Differential Geometry (Academic Press, 1966). It is
essential to have this book: the course is based on it and will not make sense without
it.
Bibliography
The set books for M201, M231 and MST 282 are referred to occasionally; they are
useful but not essential. They are:
D.L. Kreider, R.G. Kuller, D.R. Ostberg and F.W. Perkins: An Introduction to
Linear Analysis (AddisonWesley, 1966).
E.D. Nering: Linear Algebra and Matrix Theory (John Wiley, 1970).
M. Spivak: Calculus, paperback edition (W.A. Benjamin/ AddisionWesley, 1973).
R.C. Smith and P. Smith: Mechanics, SI Edition (John Wiley, 1972).
Conventions
Before starting work on this text, please read M334 Part Zero. Consult the Errata
List and the Stop Press and make any necessary alterations for this chapter in the set
book.
Unreferenced pages and sections denote the set book. Otherwise
O'Neill denotes the set book;
Text denotes the correspondence text;
KKOP denotes An Introduction to Linear Analysis by D.L. Kreider, R.G.
Kuller, D.R. Ostberg and F.W. Perkins;
Nering denotes Linear Algebra and Matrix Theory by E.D. Nering;
Spivak denotes Calculus by M. Spivak;
Smith denotes Mechanics by R.C. Smith and P. Smith.
Reference to Open University Courses in Mathematics take the form:
Unit Ml 00 22, Linear Algebra I
Unit MST 281 10, Taylor Approximation
Unit M201 16, Euclidean Spaces I: Inner Products
Unit M231 2, Functions and Graphs
Unit MST 282 1, Some Basic Tools.
M334I.1 5
1.1 INTRODUCTION AND EUCLIDEAN SPACE
Introduction
In this section we introduce some notation that will be used throughout the course.
We review vector spaces and partial differentiation. In the additional text we give
some rules for the evaluation of partial derivatives.
READ: Introduction and Section 1.1 (pages 15).
Comments
(i) Page 1: line 1 In the correspondence text we shall sometimes write
g o f for the composite function.
(ii) Page 4: Definition 1.2 From now on we shall use x, y, z in this way
only. We shall not use x as a typical element of the real line, or indeed any
set as O'Neill does on pages 1 and 2 of the Introduction.
Additional Text
Partial differentiation Suppose p = (p 1} p 2 , P3) is a point of E 3 and f : E 3 ► R
is a differentiable function. To obtain (3f/dx 1 )(p) we first construct the function
F : R >R such that F(t) = f(t, p 2 ,p 3 ) and then (3f/8x 1 )(p) =F'(p 1 ). That is we
consider x 2 and x 3 as constants and differentiate in the usual way. Partial differen
tiation satisfies the following rules. Let a, b be real numbers; f, g differentiable func
tions from E 3 to R and h a differentiable function from R to R, then
(1) a <" fH *> = , Jl > b JS (fori=l,2,3,...)
9xj 3xj 3xj
(2) ^SUiLg + f^L (fori 1,2,8,...)
9xj 3xj 9xj
(3) 3(h(f)) = h , (f? 3f (fori= 1,2,3,...)
dxj 3xj
Equation (3) may be more recognizable if the composite functions are written as
h o f and h' o f. AH these results follow from the similar rules for ordinary differen
tiation. Lists of useful derivatives are to be found in
M231 Handbook, page 45 et seq.
MST 282 Handbook, page 27.
Example Find ^x + xz sin(x 2 y)) .
3x
We have
3(2x + xzsin(x 2 y)) = 2 3x + 3(xz sin(x 2 y)) by result (1)
3x 3x 3x
2 + W sin(x 2 y) + xz 3 < sin < X 'y )) . by result (2),
3x 3x
= 2 + zsin(x 2 y) + xz sin'(x 2 y) liLZ/, by result (3)
3x
= 2 + zsin(x 2 y) + 2x 2 yz cos(x 2 y).
M334I.1
Summary
Notation
g(f)
go f
f 1
E 3
P
(Pl>P2,P3)
p + q
(Pi»P2,P3) + (qi,q2,q3)
ap
a (Pl>P2>P3)
x, y, z or x 1? x 2 , x 3
af 3f , 3f
_ , and
dx dy dz
f+g
E 2
E 1 =R
Definitions
(i) Euclidean 3space E 3
(ii) Sum of points of E 3
(iii) Scalar product of a point of E 3
(iv) Euclidean plane E 2
(v) Real line E 1 = R
(vi) Natural coordinate functions x, y, z or x t , x 2 , x 3
(vii) Pointwise addition and multiplication of functions
(viii) Composite of functions g(f)
(ix) Differentiable function
(x) Inverse function f 1
Results
Partial differentiation satisfies the following rules.
(1)
(2)
(3)
3(af+bg) = q 3f_ + b 3g_
 a
bx[
3xj 9xj
a(fg) = 9f f
3xj 9xj
+ f
ag
9xj
a(h(0) = hV) 5L
bx[ 9xj
where a, b are real numbers, f, g : E'
(i= 1,2,3,...)
(i=l,2,3,...)
(i= 1,2,3,...)
>Randh:R ► R.
Page 1 , line  4
Text, page 5
Page 2, line 16
Page 3, Definition 1.1
Page 3, Definition 1.1
Page 3, Definition 1.1
Page 3, line 3
Page 3, line  3
Page 3, line  1
Page 3, line 1
Page 4, Definition 1.2
Page 4, line 24
Page 4, line 5
Page 4, line  5
Page 5, line 22
Page 5, line 23
Page 3, Definition 1.1
Page 3, line 5
Page 3, line  2
Page 5, line 22
Page 5, line 23
Page 4, Definition 1.2
Page 4, line  5
Page 1 , line  4
Page 4, Definition 1.3
Page 2, line 16
Text: page 5
Techniques
(i) Evaluation of a function.
(ii) Simplification of a function.
(iii) Partial differentiation, using results (1), (2), (3).
M334I.1
Exercises
Technique (i)
1. Page 5, Exercises 2(b) and 2(c).
Technique (it)
2. Page 5, Exercise 1(a).
3. Page 6, Exercise 4(a). Just simplify this function.
Technique (Hi)
4. Page 5, Exercises 1(c) and 1(d).
5. Page 6, Exercises 3(a) and 3(b).
6. Page 6, Exercise 4(a).
Solutions
1. Page 5, Exercises 2(b) and 2(c).
If p is a point of E 3 , then by the pointwise definition of addition and
multiplication of functions
f(p) = ((*(p)) 2 y(p»  ((y(p)) 2 *(p)).
By Definition 1.2
x((3,l,i)) = 3, y((S,l,i))=l and z((3, 1, *)) =i
Hence
f((3,l,i)) = (3 2 .(l))(("l) 2 T)=9i.
As in the above result
f((a, 1, 1a)) = a 2 .l  l 2 .(la) = a 2 + a  1, for any real number a.
2. Page 5, Exercise 1(a).
If f = x 2 y then, since f is defined by pointwise multiplication, we have
f(p) = x 2 (p) y(p) = p\ p 2 , where p = (p u p 2 , p 3 ) is a point of E 3 .
The function g = y sin z is defined in terms of composition and pointwise
multiplication and so
g(p) = y(p) sin ( Z (P)) = P2 sin p 3 .
Hence
(fg 2 )(p) = f(p)(g(p)) 2
= P?P2 (P2 sinp 3 ) 2
P!P 2 sin 2 p 3
M334I.1
and
fg 2 = x 2 y 3 sin 2 z.
In this way we see that pointwise addition and multiplication behave exactly
as expected.
If f = x 2 y and g = y sin z then
fg 2 = x 2 y(y sin z) 2 = x 2 y 3 sin 2 z.
3. Page 6, Exercise 4(a).
If h = x 2  yz then, by the pointwise definitions of addition and multiplica
tion of functions
h (p) = Pi " P2P3> for any point p = (p 1? p 2 , p 3 ) in E 3 .
Since f(p) = h(g! (p), g 2 (p), g 3 (p)) it follows that
f(p) = (gi(p)) 2 (g 2 (p))(g3(p)).
Hence f = g?  g 2 g 3 .
In this exercise gi = x + y, g 2 = y 2 , g 3 = x + z and so
f=(x + y) 2  y 2 (x + z)
= x 2 + 2xy + y 2  xy 2  y 2 z.
4. Page 5, Exercises 1(c) and 1(d).
Now fg = x 2 y 2 sin z.
Treating x and y as constants we find that
3(fg) 2 2
_l_2i = x z y z cos z.
3z
Treating x and z as constants we find that
a 2 (fg)
9y9z
= 2x y cos z.
Now sin f = sin x 2 y.
Treating x as a constant we find that
9(sin f) 2 2
— i 4_ = x cos x y.
by
Page 6, Exercises 3(a) and 3(b).
If f = x sin (xy) + y cos (xz) then, treating y and z as constants,
7\f
= xy cos (xy)  yz sin (xz).
9x
M334I.1 9
To deal with Exercise 3(b) we need to extend result (3) to cover the partial
differentiation of the composite of three functions, h from E 3 to R and
hi, h 2 from R to itself. In this exercise it will be more convenient to use the
symbol o for the composite of functions.
3 < h * < h ' oh » = (hi . (h, . h)) 8 < h ' ° h > . by result (3),
3x 3x
= (hi o (h t o h)). {h\ o h) j^l, by result (3).
3x
Here h = x 2 + y 2 + z 2 , hj = exp, h 2 = sin
and f = sin o (exp o ( x 2 + y 2 + z 2 )) = h 2 o (hi o h). Hence
11= (sin' o (exp o (x 2 + y 2 + z 2 )). exp' o (x 2 + y 2 + z 2 ). 8 (* 2 + Y* + z *)
dx 3x
, (x 2 +y 2 +z 2 ), (x 2 +y 2 +z 2 )
= cos(e v } ') e v 7 '2x
_ g (x 2 +y 2 +z 2 ) , (x 2 +y 2 +z 2 ),
= 2x e v } ' cos (e v 7 ').
6. Page 6, Exercise 4(a).
From Exercise 3
f = x 2 + 2xy + y 2  xy 2  y 2 z
hence
— = 2x + 2yy 2 .
dx
10 M334 1 .2
1.2 TANGENT VECTORS
Introduction
In this section we review arrows. They will be called tangent vectors. At each point
of Euclidean space they form a vector space. We also reviewvector fields. They are
functions assigning to each point of Euclidean space a tangent vector based at that
point. We generalize the notion of basis and coordinates to deal with vector fields.
READ: Section 1.2 (pages 610).
Comments
(i) Page 9: Definition 2.4 Whereas a vector field V assigns one tangent
vector to each point of Euclidean space, the natural frame field assigns three.
(ii) Page 10: end of Section Since any vector field V can be expressed as
V = SvjUj, for suitable real valued functions v l5 v 2 and v 3 the natural frame
field does serve as a 'basis'. When we used the term basis for a real vector
space it implied that every vector could be expressed as a linear combination
of the basis vectors. The values of the vector fields Ui, U 2 , U 3 are linearly
independent at each point. However, not every vector field can be expressed
as a linear combination of the Uj's with constant coefficients. That is the
coefficients in the sum Ev^U^ are not necessarily real numbers. Now they are
real valued functions and the collection of real valued functions is not even a
field since it is not possible to find a multiplicative inverse of a function
which is zero at some points even if it is not always zero. These functions
form a ring and the vector fields are a finitedimensional module over this
ring.
Supplementary Comment
(i) Page 10: lines 3 and 4 These are generalizations of the usual rules for the
addition and scalar multiplication of vectors expressed in terms of co
ordinates.
The first equation is obtained as follows:
(2(v i + w i )U i )(p) = 2(vi + wi)(p)U i (p), by repeated use of the
pointwise formulas for
addition and scalar multi
plication of vector fields,
= 2(vj(p) + wi(p))Ui(p), by the definition of point
wise addition of two
functions,
= 2vj(p)Ui(p) + 2wi(p)Ui(p), by the usual rules for a
vector space,
= (SviUiXp) + (2wiUi)(p)
= (SvjUj + Zw}Uj)(p), by the pointwise rules for
the "arithmetic" of
vector fields.
Since vector fields are defined as functions on E 3 , whenever two vector
fields agree at each point they must be the same vector field.
M334 1 .2
Hence
2(vi + wi)Ui = SvjUi + wiUj.
The result
2(fv i )U i = f(2:v i U i )
follows similarly.
11
Summary
Notation
V P
T p (E 3 )
Vp + Wp
av p
V
V + W
tv
{U!,U 2 ,U 3 }
SviUi
Page 6, Definition 2.1
Page 7, Definition 2.2
Page 7, line 9
Page 7, line 9
Page 8, Definition 2.3
Page 8, line 8
Page 8, line 3
Page 9, Definition 2.4
Page 9, line 8
Definitions
i)
")
iii)
iv)
v)
vi)
vii)
Tangent vector v p
Vector part of tangent vector
Point of application of tangent vector
Parallel tangent vectors
Tangent space T p (E 3 )
Addition and scalar multiplication of tangent vectors
Vector field V
viii) Pointwise principle
ix) Addition of vector fields
x) Multiplication of a vector field by a realvalued
function
xi) Natural frame field {U^ U 2 , U 3 }
xii) Euclidean coordinate functions
xiii) Differentiable vector field
Page 6,
Definition 2.1
Page 6,
Definition 2.1
Page 6,
Definition 2.1
Page 6,
line 5
Page 7,
Definition 2.2
Page 7,
line 9
Page 8,
Definition 2.3
Page 8,
line 7
Page 8,
line 10
Page 8,
line 3
Page 9,
Definition 2.4
Page 9,
Lemma 2.5
Page 10, line 6
12 M334 1.2
Results
(i) Tp(E 3 ) is linearly isomorphic to E 3 . Page 7, line 3
(ii) Each vector field V can be uniquely expressed as
V = SvjUj, where v l5 v 2 , v 3 are realvalued functions. Page 9, Lemma 2.5
(iii) Each tangent vector (a 1} a 2 , a 3 ) p can be expressed
in terms of the natural frame field as 2ajUi(p). Page 9, line 5
(iv) Addition and multiplication by a function are
expressed in terms of coordinates by
SviUj + ZwiUi = 2(vj + wi)Uj
f(SviUi) = S(fvi)Ui Page 10, lines 3 and 4
Techniques
(i) Representation of tangent vectors in standard form
Vp. Page 6, Definition 2.1
(ii) Addition and scalar multiplication of tangent
vectors. Page 7, line 9
(iii) Evaluation of a vector field at a point. Page 8, Definition 2.3
(iv) Addition and multiplication, by realvalued
functions, of vector fields. Page 8, line 8 and
Page 9, line 2
(v) Pictorial representation of tangent vectors and
vector fields. Page 7, Fig. 1.1
(vi) Representation of tangent vectors and vector fields
in terms of the natural frame field and Euclidean
coordinates. Page 9, Lemma 2.5
(vii) Addition and multiplication, by a realvalued
function, of vector fields expressed in terms of
Euclidean coordinates. Page 10, lines 3 and 4
Exercises
Techniques (ii), (vi) and (v)
1. Page 10, Exercise 1.
Techniques (iii) and (vii)
2. Page 10, Exercise 2.
Techniques (vii), (vi), (iv) and (i)
3. Page 10, Exercise 3.
M334 1.2
Solutions
1. Page 10, Exercise 1.
(a) 3v 2w = 3(2, 1,1) 2(0,1,3)
= (6,3,3) (0,2,6)
= (6,1,9).
By the definition of addition and scalar multiplication of tangent vectors,
3v p  2w p = (3v  2w) p = (6, 1, 9) p .
Using the identity (a 2 , a 2 , a 3 ) p = SajU^p), we find that
3v p  2w p = eU^p) + U 2 (p)  9U 3 (p).
(b) Vp represents the arrow from the point p to the point p + v.
Herep = (l, l,0),v = (2, 1, 1) and hence p + v = (1, 2, 1).
Similarly, w p is the arrow from p = (1, 1, 0) to p + w = (1, 2, 3).
Now
13
2v p = (2v)p = (4,2,2)p
and
v p + w p = (v + w) p = (2, 2, 2) p
and hence they are the arrows from (1, 1, 0) to (5, 1, 2) and (1,3, 2).
^^^(1,3,2)
(5,1,2)
14 M334 1.2
2. Page 10, Exercise 2.
W  xV = 2x 2 U 2  U 3  x(xUi + yU 2 )
= 2x 2 U 2  U 3  x 2 U! xyU 2
= x 2 U 1 + (2x 2 xy)U 2 U 3 ,
by result (iv) on the addition and multiplication of vector fields in terms of
coordinates.
If p = (1, 0, 2) then x(p) = 1, y(p) = and z(p) = 2.
By the definition of the natural frame field
U^p) = (1,0, 0)p, U 2 (p) = (0, 1, 0) p , U 3 (p) = (0, 0, l) p . Hence by the pointwise
definition of adaition and multiplication
(W  xV)((l, 0, 2)) = (1) 2 (1, 0, 0) p + (2.(l) 2  (1).0)(0, 1, 0) p  (0, 0, l) p
= (l,2,l) p .
3. Page 10, Exercise 3.
(a) Rearranging and dividing by 7 we obtain
7 7
(b) V(p) = ( Pl ,p 3 p 1 ,0) p
= (xj(p), x 3 (p)  x^p), 0) p , using the definition of the
coordinate functions,
= (x^p), (x 3  X!)(p), 0) p , by the definition of linear
combinations of func
tions,
= x 1 (p)U 1 (p) + (x 3  Xi)(p)U 2 (p), by the standard identity
(a 1 ,a 2 ,a 3 ) p = 2:aiUi(p).
Hence, using the definition of linear combinations of vector fields,
V = x 1 U 1 + (x 3 x 1 )U 2 .
Alternatively we could write this as
V = xU! + (z x)U 2 .
(c) V=2(xU 1 + yU 2 )x(U 1 y 2 U 3 )
= 2xUj + 2yU 2  xUi + xy 2 U 3 , by result (iii) on addition
and multiplication in
terms of coordinates,
= (2x  x)U! + 2yU 2 + xy 2 U 3
= xU 1 + 2yU 2 + xy 2 U 3 .
(d) If p = (pi, p 2 , p 3 ) then V(p) is the vector from (p t , p 2 , p 3 )
to(l+ Pl , p 2 p 3 , p 2 ).
M334 1.2 15
Hence V(p) = ((1 + Pl , p 2 p 3 , p 2 )  (p„ p 2 , p 3 )) p
= ( 1 »P2P3~ P2»P2~ P3)p
By arguments similar to those in part (b) we find that
V = Uj + x 2 (x 3  1)U 2 + (x 2  x 3 )U 3
or alternatively
V = U 1 + y(zl)U 2 +(yz)U 3 .
(e) The vector from p to the origin is  p = p so
V (P) = (P)p = H>i> "P2, P 3 )p
Hence V =  x^  x 2 U 2  x 3 U 3 or alternatively
=  xU!  yU 2  zU 3 .
16 M334 1.3
1.3 DIRECTIONAL DERIVATIVES
Introduction
This section follows on from Sections 1.1 and 1.2. In Section 1.1 we reviewed partial
differentiation. We recalled that for any differentiable function f on E 3 and for any
point p in E 3
f^p)=:(f(t,P*P3))lt = p 1 
Instead of evaluating this derivative at t = p x we often 'shift' the function so that we
can evaluate the derivative at t = 0. That is
— (p) = — (%i + t, p 2 , p 3 )) It = 0
9xj dt
Using the natural frame field introduced in Section 1.2 we can write this as
~(p) =^(f(P + tu 1 ( P ))) t = .
There are similar formulas for (3f/3x 2 )(p) and (of/3x 3 )(p), so to find (8f/3xj)(p) we
find the initial rate of change of the function f in the direction Ui(p). In Unit
MST282 7, Work and Energy I, Section 7.1.2 we met a generalization of this
technique. If n was a unit direction vector we defined the directional derivative,
(3f/3n)(p), to be the initial rate of change of f in the direction n and proved that
 = nVf(p) = Sni — (p).
on i=i 3xj
In this section we generalize the above definition and result to deal with any
tangent vector v p and then any vector field V. We show that these directional
derivatives satisfy properties similar to those of ordinary derivatives.
READ: Section 1.3 (pages 1 114) omitting the proof of Lem ma 3.2.
Comments
(i) Page 12: Lemma 3.2 This is a generalization of the result mentioned in
the introduction. The directional derivative is
v p [f]=Wf( P ) = l;v i i£(p).
i=i oxj
It is proved in exactly the same way using a version of the 'chain rule'. This
will be dealt with in Section 1.7 and until then you should accept this result
without proof.
(ii) Page 12: Theorem 3.3.(3) The symbol * in this theorem and in Corollary
3.4 signifies the ordinary product of two numbers. The inner product of two
vectors will not be introduced until Chapter II.
(iii) Page 12: line 15 to line 3 The realvalued function V[f] is defined by
(V[f])(p)=V(p)[f].
M334 1.3 17
This equation is used in the proof of Corollary 3.4.
For computational purposes the identities Uj[f] = 3f/3xj (i = 1, 2, 3) will
be very important. We can obtain these results from Lemma 3.2. For
instance
3f
3f
3f
3f
Ui(p) [f] =U,0,0) D [f] =l.^L(p) + 0. (p) + 0. (p) (p).
3xj
9xo
dx.
dxj
(iv) Page 14; line 14 to line 21 This is a very important remark. For vector
fields V and W we consider combinations of the form fV + gW, where f and g
are realvalued functions. For functions f and g we consider combinations of
the form af + bg, where a and b are real numbers.
In the example
(xUi y 2 U 3 )[x 2 y + z 3 ] = xU 1 [x 2 y + z 3 ]  y 2 U 3 [x 2 y + z 3 ] ,
by Corollary 3.4.(1),
= xU 1 [x 2 y] +XUJZ 3 ] y 2 U 3 [x 2 y] y 2 U 3 [z 3 ], by Corollary 3.4.(2).
Supplementary Comment
(i) Page 12: Theorem 3.3.(1) and (2)
(1) If v = (vj, v 2 , v 3 ) and w = (w t , w 2 , w 3 ) then
(avp + bwp)[f] = (av + bw)p[f], by the definition of addition
and scalar multiplication of
tangent vectors,
= 2(avj + bwj) (p), by Lemma 1.3.2,
9xj
= a(2v i ^L(p)Ub(2w i ^L(p)
dxj
3xj
(2)
= av p [f] +bw p [f].
v p [af + bg] =S Vi  3(af + b gM (p), by Lemma 1.3.2,
^vifa^pj.b^p))
= af2v i £.(p)j + b/2v i iL(p).
Summary
Notation
v p [f]
V[f]
Page 11, Definition 3.1
Page 13, line 15
18 M334 1.3
Definitions
(i) Derivative of a function with respect to a tangent
vector v p [f] . Page 11, Definition 3.1
(ii) Derivative of a function with respect to a vector
field V[f]. Page 13, line 15
Results
(i) If v p = (vj, v 2 , v 3 ) p then v p [f] = 2vj— (p). Page 12, Lemma 3.2
(ii) Directional derivatives, with respect to tangent
vectors and vector fields, satisfy linear and
Leibnizian properties:
(a) (av p + bw p ) [f] = av p [f ] + bw p [f ] .
(b) v p [af + bg] =av p [f] +bv p [g].
( c ) v p [fg] = v p [f] g(p) + f(p)v p [g] . Page 12, Theorem 3.3
(a) (fV + gW)[h] = fV[h] + gW[h] .
(b) V[af + bg] = aV[f] + bV[g] .
(c) V[fg] = V[f] . g + fV[g] . Page 13, Corollary 3.4.
(iii) Directional derivatives with respect to components
of the natural frame field are partial derivatives
Uj[f] =— (i = 1, 2, 3). Page 13, line 10
Techniques
Evaluation of directional derivatives with respect to
tangent vectors or vector fields using:
(i) the definition, for tangent vectors Page 11, Definition 3.1
(ii) the definition, for vector fields Text, page 16
(iii) coordinates and partial derivatives Page 12, Lemma 3.2
(iv) the linear and Leibnizian properties and the
relationship between the natural frame fields and
partial derivatives. Page 12, Theorem 3.3,
Page 13, Corollary 3.4 and
Page 13, line 10.
Exercises
Technique (i)
1. Page 14, Exercise 1(c).
Technique (ii)
2. Determine V[f] where f = e x cos y and V = 21^  U 2 + 3U 3 .
M334 1.3 19
Technique (Hi)
3. Page 15, Exercise 2(c).
Technique (iv)
4. Page 15, Exercise 3(e). First prove that Vtf 2 ] = 2f V[f] .
Theory Exercises
5. Page 15, Exercise 4.
6. Page 15, Exercise 5. (HINT: Use Exercise 4.)
Solutions
Page 14, Exercise 1(c).
By Definition 1.3.1,
vp[f]=~(f(p + tv)) t = 0.
Here
p + tv = (2, 0, 1) + t(2, 1, 3) = (2 + 2t, t, 1 + 3t)
and
f(p + tv) = (e x cosy) (2 + 2t, t, 1 + 3t)
= e 2 + 2t cos (t), by the pointwise definition of addi
tion, multiplication and composition
of functions and the definition of the
functions x and y,
= e 2 e 2t cos t.
Hence
d
v p [f]=e 2 ^(e 2 W) t =
= e 2 (2e 2t cos t  e 2t sin t) t _ q = 2e 2 .
The value of the function V[f] at a point p is given by the formula
(v[f]Xp) = v(p)[f].
Now V(p) = 2U,(p)  U 2 (p) + 3U 3 ( P ) = (2, 1, 3) p .
Hence if p = (p t , p 2 , p 3 ) then
V( P )[f] =4(f((Pi, P 2 , Pa) + t(2, 1, 3))) t =
dt
= (f( Pl + 2t,p 2 t,p 3 +3t)) t = o
dt
_ d, p! + 2t . . Xl
~r( e cos (P2 " *)) t =
at '
= (2ePi + 2t cos(p2 . t) + ePl + 2t sin(p2 _ t )) 1 1 =
= e" 1 (2cos p 2 + sin p 2 ).
20 M334 1.3
So V[f] (p) = eP»(2cos p 2 + sin p 2 ) and hence
V[f] = e x (2cos y + sin y).
3. Page 15, Exercise 2(c).
By Lemma 1.3.2
v p [f]=Sv i JL( P ),
3xj
though in this solution we replace x lt x 2 , x 3 by x, y, z. Since f = e x cos y it
follows that
af  x 9 f x a 9f n
e A cos y, = e x sm y and = 0.
3x dy 9z
Since p = (2, 0,1)
—(p) = e\ il(p) = and il(p) = 0.
9x 8y 9z
Hence, since v = (2, 1, 3)
v p [f] = 2.e 2 + (l).O + 3.0 = 2e 2 .
4. Page 15, Exercise 3(e).
By the Leibnizian property, Corollary 1.3.4(3),
V[f 2 ] =V[f.f] =V[f].f + fV[f] =2fV[f].
Hence by the linearity property, Corollary 1.3.4(2),
V[f 2 + g 2 ]=2(fV[f]+gV[g]).
Now
V[f] =(y 2 U,xU,)[xy] =y 2 U,[xy] xU 3 [xy]
= y^lM x ^I), since UilH =— ,
9x 9z 9xj
= y 2 .y  x.O = y 3 .
Similarly,
V[g]=(y 2 U 1 xU 3 )[z 3 ]
= 3xz 2
and hence
V[f 2 + g 2 ] =2(xy.(y 3 ) + z 3 .(3xz 2 ))
= 2x(y 4  3z 5 ).
5. Page 15, Exercise 4.
Following the hint we evaluate V[x;] , where V = ZvjUj.
For fixed j we have
V[x j ]=2v i U i [x j ]=2v i ^.
9xj
M334 1.3 21
dx
Now J = 1 if i = j and is zero otherwise. Hence
dxj
V[xj] =vj for all j,
and
V = 2v i U i = SV[x i ]U i .
6. Page 15, Exercise 5.
Since V[f] = W[f] for every function f on E 3 it follows that
V[ Xi ]=W[xi] (i = l,2,3).
Hence by the result of Exercise 4
V = SV[xi] Ui = 2W[ Xi ] Ui = W.
22
M334 1.4
1.4 CURVES IN E 3
Introduction
This section follows on from Sections 1.2 and 1.3. We review two ways of defining
curves, the definition of the velocity of a curve, and show how the rate of change of
a function along a curve is related to the directional derivative with respect to the
velocity vector.
READ: Section I. 4 (pages 1521) omitting the proof of Lemma 4.6.
Comments
(i) Section 1.4 From Definition 4.1 you might think that curves always
have routes that appear in some sense smooth. However, curves can be
degenerate in several different ways. For instance, the simplest curve is the
constant curve a : 1 1 ► 0, which has as its route a single point. The curve
01 : t ' — > (sin t, 0, 0) has as its route a closed interval which it traverses
infinitely often. It is even possible to have a curve that turns sharp corners.
For instance, the curve
'(eVt'.o.O)
a : 1 1 — ► \ (0, 0, 0)
has the following route
for t <
f or t =
for t >
V(0,1,0)
(1,0,0)
(0, 0, 0)
>x
This function was chosen to ensure that the Euclidean coordinate functions
are 'infinitely' differentiable. In order for the curve to stand still, reverse or
turn a sharp corner, as in the above examples, it is necessary for the velocity
vector to take the value zero. Hence, if we restrict attention to regular curves
all these pathologies can be avoided. The definition of a 'Curve' ensures that
we can always find a regular parametrization of each component. We use the
word "closed" to describe components of a Curve that are like a distorted
circle and so not surprisingly they have periodic parametrizations. Any
component that is not like a distorted circle is like a distorted copy of the
real line and so not surprisingly has a onetoone parametrization. The circle
and a branch of the hyperbola, described on page 21, are typical of closed
and not closed Curves.
M334 1.4
23
Page 15: line 9 of the section Here the Euclidean coordinate functions
are realvalued functions on some interval I. When we described a vector field
in terms of its Euclidean coordinate functions they were realvalued func
tions on E 3 .
Supplementary Comments
(i) Page 16: Example 4.2(3)
a(t) = (2cos 2 t, sin 2t, 2sin t) = (2cos 2 1, 2sin t cos t, 2sin t).
Since
(2cos 2 t) 2 + (2sin t cos t) 2 + (2sin t) 2
= 4cos 2 t(cos 2 t + sin 2 1) + 4sin 2 t = 4(cos 2 t + sin 2 t) = 4
it follows that a(t) belongs to S.
Also
a(t)  (1,0,0) = (2cos 2 t  1, sin 2t, 2sin t) = (cos 2t, sin 2t, 2sin t) and since
cos 2 2t + sin 2 2t = 1 it follows that a(t) belongs to C.
As t tends to the point a(t) tends to (2,0,0), and as t tends to tt/2 the point
a(t) tends to (0,0,2), and hence ex. follows the route sliced from C by the
sphere S.
(ii) Page 18: line 6 Since cq(t) = pj + qj (i = 1, 2, 3) it follows that
dai/dt(t) = qj (i = 1, 2, 3) and hence oc'(t) = (q lf q 2 , q 3 )a(t)
(iii) Page 19: Lemma 4.5 By the definition of the composite of two
functions,
(<*(h))(s) = a(h(s)) and («i(h))(s) = <*i(h(s)).
Hence, since j3(s) = a(h(s)) = (a(h))(s),
0'(s) = (a(h))'(s).
Since a(h(s)) = (a 1 (h(s)), a 2 (h(s)), a 3 (h(s))),
(«(h))(s) = ((a,(h))(s), (a 2 (h))(s), (a 3 (h))(s))
and by Definition 4.3
(a(h))'(s) = ((«i(h)) f (s), (a 2 (h))'(s), (a 3 (h))'(s)).
The result now follows as on page 19.
Summary
Notation
<*(*) = Wt),a 2 (t),a 3 (t))
a = (<*i, <*i, « 3 )
OL
«'(t)
C: f=a
Page 15, line 7
Page 15, line 10
Page 15, Definition 4.1
Page 17, Definition 4.3
Page 20, line 5.
24
Definitions
(i) Euclidean coordinate functions of a curve
(ii) Curve a.
(iii) Velocity vector of a curve a'(t)
(iv) Reparametrization of a curve
(v) Periodic curve
(vi) Period of a curve
(vii) Regular curve
(viii) 'Curves' on the plane defined implicitly C: f = a
(ix) Closed curve on the plane
M334 1.4
Page 15, line 9
Page 15, Definition 4.1
Page 17, Definition 4.3
Page 18, Definition 4.4
Page 20, line 13
Page 20, line 15
Page 20, line 10
Page 20, line 5
Page 21, line 5 and
Text: page 22
Examples
(i) Straight line. a(t) = p + tq
(ii) Helix. a(t) = (a cos t, a sin t, bt)
Page 15, Example 4.2.(1)
Page 15, Example 4.2.(1)
Results
(i) If ]8 is the reparametrization of a by h, then
^(s)=ffl(s).«'(h(s)).
Ids/
(ii) If ex is a curve in E 3 and f is a differentiable
function on E 3 , then
„. (t)[f] =M (t)
Page 19, Lemma 4.5
Page 19, Lemma 4.6
Techniques
(i) Determination of a parametric representation of a
straight line.
(ii) Determination of the velocity vectors of a curve.
(iii) Determination of the effect of reparametrization
on velocity vectors.
(iv) Calculation of the rate of change of a function
along a curve by means of the directional deriva
tive with respect to the velocity vector.
(v) Parametrization of implicitly defined 'Curves' in
E 2 .
Page 15, Example 4.2.(1)
Page 17, Definition 4.3
Page 19, Lemma 4.5
Page 19, Lemma 4.6
Page 21.
M334 1.4 25
Exercises
Technique (i)
1. Page 21, Exercise 5.
Technique (ii)
2. Page 21, Exercise 1.
Technique (Hi)
3. Page 21, Exercise 3. Evaluate <*'(7r/4), 0'(l/v/2) and (dh/ds)(l/ v /2). Verify
Lemma 1.4.5 in this case.
l
(Note: sin" 1 was called arc sin in M231 and (sin" 1 )' (s) = 1/(1  s 2 ) 2 ".)
Technique (iv)
4. Page 21, Exercise 7.
Technique (v)
5. Page 22, Exercise 10.
Solutions
1. Page 21, Exercise 5.
Suppose the line is given by a(t) = p + tq. If it passes through the points Pi
and p 2 we can assume a(0) = pj and a(l) = p 2 . Then p = Pi and q = p 2  pi
and hence a(t) = Pi + t(p 2  Pi). If Pi = (1,3, 1) and p 2 = (6, 2, 1) the curve
is
a(t) = (1, 3, 1) + t((6, 2, 1)  (1, 3, 1))
= (l,3,l) + t(5,5,2)
= (l + 5t, 3 + 5t,l+2t).
Similarly the line which passes through (1, 1,0) and (5, 1,1) is given by
0(s) = (1  4s, 1  2s, s).
If the lines meet, there are real numbers t and s such that
«(t)=/3(s).
i.e. 1 + 5t = 1  4s
3 + 5t = 1  2s
1 + 2t = s.
The first two equations can be solved uniquely giving t = 2, s = 3, which is
also a solution of the third equation, so the lines meet at
a(2)=0(3) = (ll,7,3).
26 M334 1.4
2. Page 21, Exercise 1.
a(t) = (2cos 2 t, sin2t, 2sint).
Hence by Definition 1.4.3,
a'(t) = (£(2cos 2 1), _£_(sin 2t), jL(2sin t)\
Ut dt dt Mt)
= (4cos t sin t, 2cos 2t, 2cos t) ( 2cc .s 2 1, sin 2t, 2sin t)
and
""Q' ("2. °V2) (1,1,^2)
3. Page 21, Exercise 3.
By Definition 1.4.4 the reparametrized curve is given by
3(s) = a(h(s)) = a(sin _1 s).
Now a(t) = (2cos 2 t, sin 2t, 2sin t) and to make progress we express a(t) as a
function of sin t.
We find that a(t) = (2(1  sin 2 t), 2sin t (1  sin 2 t)*,2sin t), using the identi
ties cos 2 t + sin 2 t = 1 and sin 2t = 2sin t cos t.
Hence 8(s) = o^sin" 1 ^))
= (2(ls 2 ),2s(ls 2 )t2s).
The function sin" 1 has domain [1, 1] and so is well defined for < s < 1
and takes values in the interval < s < 7r/2.
Now
a'(t) = (4cos t sin t, 2cos 2t, 2cos t)
and
«'I1\=U.±.L, 0, 2._L) = (2,0,V2).
U/ \ n/2^2 y/21
Also
^(s) = (4s,2(ls 2 f_i!L 1 ,2)
\ (1s 2 ) 2 /
and
/3'[_L1 = (2V2, 0, 2).
"P
Finally
%)L^ and iHlW
ds (1s 2 ) 2 dsV2/
1 \ _ • i / 1 \ it
Hence
j3'[ 1 = [ ] • <*'(— ] and, since h ( x ] = sin
V2/ ds\V2/ U/ V2/ \V2/ 4
this agrees with Lemma 1.4.5.
4. Page 21, Exercise 7.
Let a(t) = (t, 1 + t 2 , t),
0(t) = (sint, cost, t),
7(t) = (sinht, cosht, t).
M334 1.4 ' 27
Then a'(0) = 0'(O) = 7'(0) = (1, 0, 1) (q, 1, 0) The curves have the same
initial velocity v p .
By Theorem 1.4.6
v p [f] =l(f(«))t = =W))\t = = (f(7))t =
dt dt dt
and since f = x 2  y 2 + z 2 we have
_£(f(a)) t=: o=i(t 2 (l + t 2 ) 2 + t 2 ) t = o = i(lt 4 ) t = o=0,
dt dt dt
(f(»)t = = l(sin 2 t  cos 2 t + t 2 ) t = o=0,
dt dt
l(f(T))t = = i(sinh 2 t  cosh 2 t + t 2 ) t = = 0.
dt dt
This is the answer we expect since by Lemma 1.3.2
(i. o. i) (0, 1, 0)M = (i— + o.ii + i.W 1, 0))
\ 3x 3y 3z/
= 2x((0,l,0)) = 0.
5. Page 22, Exercise 10.
(a) a : 1 1 — ► (•§ cos t, sin t)
28
M334 1.4
(b) « : t h— t>
l3t
(c) art i — ►(t, e 1 )
M334 1.4
(d)
a: ti — ►
0<t<l
(t, (1  t*)*)
29
>x
30 M334 1.5
1.5 1 FORMS
Introduction
This section follows on from Section 1.3, Directional Derivatives.
In this section we introduce 1forms. A 1form is an extension of the idea of linear
functional. A linear functional is a linear transformation between a vector space V
and the underlying vector space of reals R. Now we have a collection of vector
spaces, the tangent spaces T p (E 3 ), and we require our 1forms to map any tangent
vector to a real number in such a way that the restriction to any one tangent space is
a linear functional.
READ: Section I. 5 (pages 2225).
Comments
(i) Page 23: Definition 5.2 If V is a vector field, then the definition of the
differential implies that df(V) = V[f]. This arises from the pointwise
definition in the following way. For any point p in E 3
(df(V))( P ) = df(V( P )) = V(p)[f] = (V[f] )(p)
and so df(V) =V[f] .
(ii) Page 23: Example 5.3.(1) The symbol 5jj is a very useful tool and
identities such as
SvjSy = Vl 8 h + v 2 5 i2 + v 3 5 i3 = Vi (i = 1, 2, 3)
J
occur frequently. To check this identity we look at an example, say i = 2.
Then
2 vj5 2J  = Vl 5 21 + v 2 S 22 + v 3 5 23
J
= Vj .0 + v 2 .l + v 3 .0 = v 2 .
(iii) Page 23: line 1 We use the same symbol for a real number and for the
constant function taking that value. Hence we can write dxj (Uj) = fa. We
obtain this result from first principles as follows
dx i (U j ) = U j [x i ]=^i = 5 ij .
3 Xj
This is the basic result to remember. It can be used to obtain the results of
both Example (1) and (2) as follows:
(1) dxi(vp) = dxi/svjUjtp)! = 2v j dx i (U j (p))
= ZvjSij = Vi .
J
(2)
*(vp) = fcfidxij (evjUjCp) 1
2f i (p)v j dx i (U j (p))= Sf i (p)vj8 ij
Zfi(p)vi.
i
09
M334 1.5
31
(iv) Page 25: Lemma 5. 7 This follows from the rule for the partial differen
tiation of a composite given in the Additional Comments to Section 1.1.
Supplementary Comments
(i) Page 23: lines 5 to 10 Two functions are equal if they take the same
value at each point p of E 3 .
Now
(0(fV + gW))(p) = 0((fV + gW)(p)), by the definition of the
evaluation of a 1form on a
vector field,
by the pointwise definition
of operations on vector
fields
= 0(f(p)V(p)+g(p)W(p)),
= f(p)0(V(p))+g(p)0(W(p)),
= f(p)Wv))( P ) + g(p)(0(w))(p)
= (f0(V) + g0(W))(p).
Hence
0(fV + gW) = f 0(V) + g 0(W).
Similarly, since
((f0 + g 0)(V))(p) = (f0 + g 0)(V(p))
= f(p)0(V(p))f g (p)0(V(p))
= f(p)(«v))(p) + g(p)(WV))(p)
= (f0(V) + g0(V))(p),
it follows that (f0 + g0)(V) = f0(V) + g0(V).
by the definition of a
1form
(ii) Page 23: Definition 5.2 The function df is a 1form since
df(av p + bw p ) = (av p + bw p )[f]
= av p [f] +bw p [f], by Theorem 1.3.3.(1),
= adf(vp) + bdf(w p ).
(iii) Page 24: the proof of Lemma 5.4 Since the natural frame field forms a
basis, when restricted to each vector space, any 1form that is zero on all
three of these vector fields must be the zero 1form. For j = 1, 2, 3,
 20(Ui)d Xi (Uj) = 0(Uj)  SflUiJdxifUj)
= 0(Uj)  ZflUOSij = 0(Uj)  0(Uj) = 0.
Hence  20(Ui)dxi = and = 20(U i )dx i .
32 M334 1.5
(iv) Page 24: the proof of Corollary 5.5 This can be proved directly using
Lemma 1.5.4.
df = 2df(Ui)dxi, by Lemma 1.5.4,
i
= 2Uj[f] dxi, by Definition 1.5.2,
i
= 2 dxj.
idxj
(v) Page 25: the worked example How do we obtain the first term 2xy dx?
Now f = x 2 y + . . . and hence df = d(x 2 y) + • • • = <*(x 2 )y + x 2 dy + . . . , by
Lemma 1.5.6. To evaluate d(x 2 ) rigorously we need to introduce the
squaring function S : 1 1 — ► t 2 for which S' : 1 1 — ► 2t.
Then d(x 2 ) = d(S(x)) = S'(x)dx, by Lemma 1.5.7,
= 2x dx.
Hence df = 2xy dx + x 2 dy + . . .
Finally v p [f] = df(v p ) is evaluated using the result of Example 5.3.(2).
Summary
Notation
«Vp)
0p
f0
0(V)
df
df(v p )
df(V)
dxj
Page 22, Definition 5.1
Page 22, Definition 5.1
Page 22, line 11
Page 22, line 7
Page 22, line 4
Page 22, line 2
Page 23, Definition 5.2
Page 23, Definition 5.2
Text: page 30
Page 23, Example 5.3.(1)
Page 23, line 12
Definitions
(i) 1form
(ii) Addition of 1 forms
(iii) Multiplication of a 1form by a realvalued
function
(iv) The effect of a 1form on a vector field
(v) The differential of a real valued function df
(vi) Kronecker delta 5jj
Page 22, Definition 5.2
Page 22, line 7
Page 22, line 4
Page 22, line 2
Page 22, Definition 5.2
Page 23, line 12
M334 1.5 33
Results
(i) 0(V) is linear in both the 1 forms and vector
fields V. Text: page 31
(ii) dxi(vp) = vj. Page 23, Example 5.3.(1)
(iii) If = Zfidxi, then 0(v p ) = S^p)^. Page 23, Example 5.3.(2)
(iv) dxj(Uj) = 5jj. Page 23, line 1
(v) = 20(Ui)dxj. Page 24, Lemma 5.4
(vi) df = 2 dxj. Page 24, Corollary 5.5
9xj
(vii) d(f+g) = df+dg. Page 24, line 9
(viii) d(fg) = gdf + fdg. Page 24, Lemma 5.6
(ix) If f : E 3 ► R and h : R >R then d(h(f)) = h'(f)df. Page 25, Lemma 5.7
Techniques
(i) Recognition of a 1form, by comparison with the
formula 0(av p + bw p ) = a0(v p ) + b0(w p ). Page 22, Definition 5. 1
(ii) Expressing a 1form in standard form using the
formula = 20(Ui)dxi. Page 24, Lemma 5.4
(iii) Evaluation of the effect of a 1form on a tangent
vector using the formula
0( v p) = 2fi(p)vi where = S^dx;. Page 23, Example 5.3.(2)
i i
(iv) Evaluation of the effect of a 1form on a vector
field using the identity dxi(Uj) = 5jj and linearity. Text: page 30
(v) Calculation of differentials using the 'chain rule'
At 9f a Jf j 9f J
df ~ dx i + dx 2 + dx 3 . Page 24, Corollary 5.5
3xj dx 2 3x 3
(vi) Calculation of differentials using the linear,
Leibnizian and composite properties
d(f + g) = df + dg, Page 24, line 9
d ( f g) = g df + f dg, Page 24, Lemma 5.6
d(h(f)) = h'(f)df. Page 25, Lemma 5.7
(vii) Evaluation of directional derivatives using
differentials
v p[ f ] = d f(v p ). Page 23, Definition 5.2
Exercises
Techniques (i) and (ii)
1. Page 26, Exercise 7.
34 M334 1.5
Technique (Hi) l
2. Page 25, Exercise 1(c).
Technique (iv)
3. Page 25, Exercise 3. Evaluate on W only.
Technique (v)
4. Page 26, Exercise 5(a).
Technique (vi)
5. Page 26, Exercise 5(b). [Note: (tan" 1 )' (t) =
ix' 1
\ 1+V
Technique (mi)
6. Page 26, Exercise 6(b).
Solutions
Page 26, Exercise 7.
To be a 1form, needs to be linear on each tangent space Tp(E 3 ). That is,
the mapping
(vi, v 2 , v 3 ) i — ► ((vi, v 2 , v 3 )p) must be linear for each p in E 3 .
(a) The mapping is
(Vi, V 2 , V 3 ) I ► Vj  v 3 ,
which is a linear mapping from E 3 to R independent of whichever
point p we choose. Hence this does give us a 1form.
Now0(U 1 (p)) = 0((l,O,O) p ) = l
0(U 2 (p)) = 0((O,l,O)p) = O
0(U 3 (p)) = 0((O, 0, l) p ) = 1
and hence, since = 20(Uj)dxj,
i
= l.dx!  l.dx 3 = dxj  dx 3 .
(b) This is not linear, since, for a fixed point p withp^ps and
nonzero vector v, the definition gives
0(2v p ) = 0(v p ) = Pl p 3 ^O
while we would expect 0(2v p ) = 20(v p ) if were linear.
(c) For fixed p = (pi, p 2 > Pa) the mapping is
(vi, v 2 , v 3 )i ►PsVi + p^,
which is linear and hence is a 1form.
Now 0(U^p)) = 0((1, 0, 0) p ) = p 3 = x 3 (p) and hence 0(U t ) = x 3 ;
0(U 2 (p)) = 0((O, 1, 0) p ) = pi = Xi(p) and hence 0(U 2 ) = x i;
0(U 3 (p)) = 0((O, 0, l) p ) =
and hence = 20(U})dxj = x 3 dxj + X!dx 2 .
M334 1.5 35
(d) For fixed p the mapping v p i — ►v p [x 2 + y 2 ] is linear, by Theorem
1.3.3.(1), and since v p [x 2 + y 2 ] is a real number p is a linear func
tional and is a 1form. Now by the definition of a differential
v p [x 2 + y 2 ]=d(x 2 + y 2 )(v p ).
Hence, since 0:v p i — >v p [x 2 + y 2 ] = d(x 2 + y 2 )(v p ), it follows that
= d(x 2 + y 2 ) = 2x dx+ 2y dy.
(e) This mapping is linear for each fixed p and so does define a 1form.
Obviously it is the zero 1form, for which again we use the greatly
overworked symbol 0.
(f) This fails to five us a 1form for much the same reason as for part
<b).
2. Page 25, Exercise 1(c).
If \jf = Sfjdxi then ^( Vp ) = 2fi(p) Vi .
i
Here i// = (z 2  l)dx  dy + x 2 dz and hence
^( v p) = (Pi _1 ) v i " V 2 + P?V3.
If v p = (1, 2, 3) (0 , _ 2 , l) then tf,(v p ) = 2.
3. Page 25, Exercise 3.
0(W) = (x 2 dx  y 2 dz)((xy + yz)\J l  yzU 2  xyU 3 )
= x 2 (xy + yz)dx(Uj)  x 2 yz dx(U 2 )  x 3 y dx (U 3 )
y 2 (xy + yz)dz(U x ) + y 3 z dz(U 2 ) + xy 3 dz(U 3 ), by linearity,
= x 2 (xy + yz) + xy 3 = x^ + x 2 yz + xy 3
= xy(x 2 + xz + y 2 ),
since dx^) = dz(U 3 ) = 1 and all the other similar terms are zero.
Page 26, Exercise 5(a).
By Corollary 1.5.5, df =^£dx + JUdy +_^ldz.
3x By 9z
i
Here f = (x 2 + y 2 + z 2 ) 1 and hence ^1= x(x 2 + y 2 + z 2 j"t
3x
Using similar results for and_we obtain
3y dz
,i
df = (x 2 + y 2 + z 2 ) 2 (x dx + y dy + z dz).
5. Page 26, Exercise 5(b).
When dealing with the function tan _1 (y/x) we must restrict ourselves to a
region of E 3 on which the function x is never zero. The largest such domain
is{pEE 3 : Pl #0}.
36 M334 1.5
Now d(tan _1 (y/x)) = (tan" 1 )'(y/x)d(y/x), by Lemma 1.5.7,
1
1 + (y/x) 2
x 2 /l
•d(y/x),
* [Idy + yd (1)\ , by Lemma 1.5.6,
+ v 2 \x x /
x 2 + y
_ x 2 /dy_ ydx
x 2 + y 2 \ x x 2
x dy  y dx
x 2 + y 2
by Lemma 1.5.7,
6. Page 26, Exercise 6(b).
By Corollary 1.5.5
d(xeYZ) = ^fl!> dx + d ±^l dy + S J^1 dz
3x 3y 9z
= eV z (dx + xz dy + xy dz).
Hence df[v p ]= eP*P 3 ( Vl + Pl p 3 v 2 + Pl p 2 v 3 ) and if v p = (1, 2, 3) ( , 2, 1)
it follows that df [v p ] = e" 2 .
M334 1.6 37
1.6 DIFFERENTIAL FORMS
Introduction
This section follows on from Section 1.5, where we gave an abstract definition of a
1form and then proved that any 1form is a combination of the differentials of the
natural coordinate functions. In this section we introduce 0, 1, 2 and 3forms as
formal expressions in terms of these differentials and define the wedge product and
exterior derivative of such forms. In Chapter IV and Chapter VI we shall see that
these higher dimensional forms can be defined in terms of multilinear functionals
operating on tangent vectors.
READ: Section 1.6 (pages 2631).
Comments
(i) Page 27: lines 15 to 19 Take these formal expressions as the definitions
of 0, 1, 2, 3forms and treat the previous paragraphs as motivation only.
In certain cases it is possible to simplify the formal expression for a pform
without introducing any ambiguity.
(a) If some, but not all, of the coefficient functions are the zero func
tions we can omit the corresponding part of the linear combination.
For example
e x dx + xyz dy + Odz = e x dx + xyz dy
and
z 2 dx dy + dx dz + dy dz = z 2 dx dy.
(b) If all the coefficient functions are zero we have the zero pform and
when no confusion can arise we just write this as 0.
That is,
(i) the constant function with value is the zero 0form.
(ii) Odx + Ody + Odz = is the zero 1form.
(iii) dx dy + dx dz + dy dz = is the zero 2form.
(iv) dx dy dz = is the zero 3form.
It is very unlikely that this abuse of the symbol will ever lead to an
ambiguity.
(c) If some of the coefficient functions are the constant function with
value 1 we omit them. For example,
eY dx + 1 dy + 1 dz = eY dx + dy + dz
and eY dx dy + dx dy + 1 dy dz = eY dx dy + dy dz.
(d) If some of the coefficient functions are the constant function with
value 1 we omit the symbol "1". For example,
1 dx dy + ( 1) dx dz + Ody dz = dx dy  dx dz
and (1) dx dy dz = dx dy dz.
38 M334 1.6
More generally, if some coefficient function is f, for some more
commonly used function f, we write ... f ... instead of
. . . + (f) . . . . For example,
x dx + (y) dy + dz = x dx  y dy.
(ii) Page 27: line 23 Using the interpretation of 1 forms given in the last
section we can prove that to add 1 forms or multiply them by some real
valued function all we have to do is to add the coefficient functions or
multiply them by the given function. That is,
Sfidxj + Sgidxj = 2(fi + gi )dx i5
f(2fid Xi ) = 2(ffi)dxi.
Since we have no way of interpreting 2 and 3forms yet the best we can do
is to define addition and scalar multiplication in a similar way. That is
(f ! dx dy + gjdxdz + r^dydz) + (f 2 dxdy + g 2 dxdz + h 2 dydz)
= (fj + f 2 )dxdy + ( gl + g 2 )dxdz + (hi + h 2 )dydz;
k(fdxdy + gdxdy + hdydz) = kfdxdy + kgdxdz + khdydz;
fjdxdydz + f 2 dxdydz = (f t + f 2 )dxdydz;
g(fdxdydz) = gfdxdydz.
(iii) Page 27: Example 6.1 This example serves as a definition of the wedge
product.
The product is given by the following procedure.
(a) Expand using the distributive law, bringing the coefficient functions
to the front, dropping the symbol a but maintaining the order of the
differentials.
(b) Drop terms with a repeated differential.
(c) Reorder the strings of differentials using the alternation rule.
(iv) Page 28: Definition 6.3 The wedge products dfjAdxj are expanded using
Corollary 1.5.5.
That is,
dfjAdxj = l^h dx! + Oh. dx 2 + Oh. dx 3 ] Adxj
\dx! 3x 2 dx 3 /
= i dxjdxj + L dx 2 dxj + L dx 3 dxj.
bxi 9x 2 9x 3
The definition of the exterior derivative can be extended to cover 2forms. If
17 = f dx dy + g dx dz + h dy dz
then
&q = df Adx dy + dgAdx dz + dhAdy dz
and this expression is again expanded using Corollary 1.5.5.
M334 1.6
39
(v) Page 29: line 13 Though we can multiply 1forms by any realvalued
functions, in this formula the coefficients a and b are only real numbers or
equivalently constant functions.
(vi) Page 30: lines 11 and 12 Since dx dy dz is obtained from dz dx dy
by two interchanges the expression is multiplied by (1) 2 = 1.
Summary
Notation
fdx
+ gdy + hdz
Page 27, line 16
fdx
dy + gdxdz + hdydz Page 27, line 18
fdx(
dy dz
Page 27, line 19
0A^
Page 27, Example 6.1
d0
Page 28, Definition 6.3
Definitions
(i)
Differential form ^
0form
1form
2form
Page 27, lines 1519
3form
)
(")
Addition and multip
lication of forms Page 27, lines 2023
(iii)
(iv)
Wedge product 0a \jj
Exterior derivative d<
Page 27, Example 6.1
P Page 28, Definition 6.3 and
Text: page 38
Results
(i) The wedge product of 1forms satisfies the
alternation rule 0a \]j =^a0
(ii) The exterior derivative satisfies linear and
Leibnizian properties
d(a0 + b\^) = ad0 + bdi//
d(fg) = dfg + fdg
d(f0) = dfA0 + fd0
d(0Al//) = d0Al// — 0Ad^.
Page 28, Lemma 6.2
Page 29, line 13 and
Theorem 6.4.
40 M334 1.6
Techniques
(i) Addition and scalar multiplication of
differential forms. Page 27, lines 2023
(ii) Expansion of the wedge product of
differential forms. Page 27, Example 6.1
(iii) Expansion of the exterior derivative. Page 28, Definition 6.3
(iv) Expansion of the exterior derivatives using
the linear and Leibnizian properties. Page 29, Theorem 6.4
Exercises
Technique (i)
1. (a) Add the 2forms
xdxdy + ydxdz and xdxdy + e x Ydxdz + 2dydz.
(b) Multiply the second of the above 2forms by the function e~ x Y.
Technique (ii)
2. Form the wedge product of
(a) yzdx + dz and dy + zdz,
(b) yzdx + dz and xdxdy + e x Ydxdz + 2dydz.
Technique (iii)
3. Find the exterior derivative of
(a) xyz dx + dz,
(b) xyz dx dy.
Technique (iv)
4. Page 31, Exercise 3.
Theory Exercise
5. Page 31, Exercise 7.
Solutions
1. (a) (xdxdy + ydxdz) + (xdxdy + e x Ydxdz + 2dydz)
= (x + (x))dxdy + (y + e x Y)dxdz + 2dydz
= dx dy + (y + e x Y)dx dz + 2dy dz
= (y + e x Y)dx dz + 2dy dz.
M334 1.6 41
(b) e" x Y(xdxdy + e x Ydxdz + 2dydz)
= xe x Ydxdy + e x Ye x Ydxdz + 2e" x Ydydz
= xe'^dx dy + 1 dx dz + 2e~ x Ydy dz
= xe x Ydxdy + dxdz + 2e" x Ydydz.
2. (a) (yzdx + dz) a (dy + zdz)
= yz dx dy + yz 2 dx dz + dz dy + z dz dz, by the distributive rule,
= yz dx dy + yz 2 dx dz + dz dy, since dzdz = 0,
= yzdxdy + yz 2 dxdz  dydz, by the alternation rule,
(b) (yzdx + dz) a (xdxdy + e x Ydxdz + 2dydz)
= xyzdxdxdy + yze x Ydxdxdz + 2yzdxdydz
x dz dx dy + e x Ydz dx dz f 2dz dy dz, by the distributive law,
= 2yz dx dy dz + x dz dx dy, since dx dx ay = dx dx dz
= dz dx dz = dz dy dz = 0,
= (2yz  x)dx dy dz, since dz dx dy = dx dy dz.
*• (a) d(xyzdx + dz) = d(xyzdx + ldz)
= d(xyz) A dx + d(l) a dz
= d(xyz) A dx,
since the differential d(l), of the constant func
tion with value 1, is the zero 1form and since
a dz is the zero 2form,
= (yzdx + xzdy + xydz) a dx,
by Corollary 1.5.5,
= yzdxdx + xzdydx + xydzdx,
by the distributive rule,
= xzdydx + xydzdx,
since dx dx = 0,
= xzdxdy xydxdz,
by the alternation rule.
(b) d(xyzdxdy) = d(xyz) a dxdy
= (yzdx + xzdy + xydz) a dxdy
= yz dx dx dy + xz dy dx dy + xy dx dx dy
= xydz dxdy
= xydxdydz.
42 M334 1.6
4. Page 31, Exercise 3.
By Corollary 1.5.5, df = 2 JUdxj.
i 3xj
By Definition 1.6.3, d(df) = 2 d PI] a dxj.
i \3xi/
Again using Corollary 1.5.5 we have
d(df) = 2 1 2 32f dx; ) a dxi = 2 2 92f dxjdxj.
1 \ J a Xi 3xj / J J 3 Xi 3xj
There are nine terms in this sum, which we must show is zero. We know that
the terms involving dxidx!, dx 2 dx 2 and dx 3 dx 3 are already zero. The
remaining six terms occur in pairs such as
dx 2 dx 1 + dx!dx 2 .
dxj9x 2 dx 2 3xi
By the alternation rule this is
I + dxidx 2 .
\ 9x 2 3x! 3xj3x 2 /
But we know that 3 2 f/3x 1 dx 2 = 3 2 f/3x 2 3x! for suitably differentiable
functions, and so the remaining terms in the sum for d(df) cancel in pairs.
Hence d(df) = 0.
By Theorem 1.6.4,
d(fdg) = df a dg + fd(dg),
= df a dg, by the above result.
Page 31, Exercise 7.
Any 1form is of the form
= 2 fjdxj,
i
where f j and xj are differentiable functions.
By Definition 1.6.3,
d0 = 2 df} a dx}
i
and by Theorem 1.6.4 (3),
d(d0) = 2 (d(dfi) a dxj  df} a d(dxi)).
i
By Exercise 3 we know that both d(dfj) and d(dxj) are zero and so
d(d0) = 0.
M334 1 1.7 43
1.7 MAPPINGS
Introduction
This section follows on from Sections 1.3 and 1.4.
In Section 1.3 we introduced directional derivatives in order to differentiate func
tions from E 3 to R, and in Section 1.4 we introduced the componentwise differen
tiation of functions from R to E 3 . In this section we combine these ideas in order to
differentiate functions from E n to E m , for any positive integers n and m. The
derivative mappings of these functions turn out to be linear transformations between
corresponding tangent spaces and we shall describe the matrices representing them.
READ: Section 1.7 (pages 3239).
Comments
(l) Section 1.7 In this section there are several straightforward general
izations of definitions that have previously been applied to E 3 only. Those
that are not specifically mentioned are:
(a) Natural coordinate functions x 1} . . . , x n , which generalize those of
Definition 1.1.2. If p is a point in E n then xj(p) = pj (i = 1, 2, . . . , n).
(b) Tangent vectors v p in E n , which generalize those of Definition 1.2.1.
The tangent vector v p represents the "arrow" from p to p + v in E n .
(c) Curves a(t) = (a^t), . . . , a n (t)) in E*\ which generalize those of
Definition 1.4.1.
(d) Straight lines 1 1 — ► p + tv in E n .
(e) The velocity vector a'(t) = ((d ai /dt)(t), . . . , (da n /dt)(t)) a(t) for a
curve a in E n , which generalizes Definition 1.4.3.
(f) The directional derivative v p [f] of a function f on E n with respect to
a tangent vector v p , which generalizes that of Definition 1.3.1. That
is>v p [f] =(d/dt)(f(p + tv)) t = .
(ii) Page 36: line 3
F*(v) = F( p + tv) r i(0)
= ((Pi + tv 1 ) 2 (p 2 +tv 2 ) 2 ,2(p 1 +t Vl )(p 2 +tv 2 ))'(0)
= (2( Pl + tw l )v 1 2( Pl +tv 2 )v 2 ,2( Pl +tv,)v 2 +2 Vl (p 2 +tv 2 ))(0)
= 2(p!V!  p 2 v 2 , pjv 2 + p 2Vl ).
(iii) Page 37: following Corollary 7. 7 To obtain the jth column of a matrix
representing a linear transformation with respect to given bases, we find the
coordinates for the image of the jth basis vector in the domain, in terms of
the basis for the codomain.
44 M334 1.7
In Example 7.3(2) we take as basis vectors for the domain U^p) and U 2 (p)
and for the codomain Uj(F(p)) and U 2 (F(p)). Hence, by Corollary 7.7, the
linear transformation is represented by
lili(p) 2!i(p)\
9u 9v
t 3u dv /
where f j = u 2  v 2 and f 2 = 2uv.
All such matrices are called Jacobian matrices.
Hence at p = (pi ,p 2 ) the Jacobian matrix is
/2u(p) 2v(p)\ /2pi 2p 2 ^
\2v(p) 2u(p)/ \2 P2 2p u
The image of a typical vector v = (v x ,v 2 ) under this transformation is
which is the answer obtained on page 36, line 3. Note that while dealing with
matrices we represent tangent vectors by column vectors.
Additional Text
(a) Let f be a mapping from R to R. Then the Jacobian matrix is the
1 X 1 matrix (df/dt), where here we do not need partial derivatives since f is
a function of t only.
Now a tangent vector in R, v p is an arrow from p to p + v,
v
p + V
and this is mapped by f^p to the tangent vector v(df/dt)(p) at f(p). That is it
is multiplied in length by (df/dt) (p).
df, ,
v (p)
dt
» ► •—
f( P ) f(p)+vii( P )
dt
M334 1.7
45
(b) Suppose we have a curve a : R ► R 3 , where a has coordinate functions
(«!, a 2 , a 3 ), i.e.
a : t .— > a(t) = (^(t), a 2 (t), a 3 (t)).
Then we obtain the 3X1 Jacobian matrix
dt
da,
dt
A tangent vector v, at p, is mapped to the tangent vector a* p (v), at a(p).
Reverting to row vectors, this is the tangent vector
vi^( P ), vi^( P ), v^( P) )
dt dt dt / a(p)
(c) Suppose we have a realvalued function F : E 3 >R, then the Jacobian
matrix is the 1 X 3 matrix
/3F dF 3F\
\3x, dxj 3x 3 i
If v p  v 2 is a typical tangent vector, it is mapped by F^
to the tangent vector
/ 3F , . 3F , . 3F , A
(P)» (p)» (p) /vA at F(p) in R.
\ 3x t 3x 2 3x 3 / / \
v 2
, V 3
46 M334 1.7
This is V! (3F/3x 1 )(p) + v 2 (3F/8x 2 )(p) + v 3 (3F/3x 3 )(p), which we have
already met as the directional derivative v p [F] .
p + v
 » 3F . ., 3F , x , 3F , ,
1 V 1T" (P) +v 2— (P) +V 3T (P)
OXj ox 2 ox 3
F(p) F(p)+v.VF(p)
(d) The Composite Rule Suppose we have two mappings F : E n ► E m and
G : E m ► E r , then we can form the composite mapping G ° F : E n ► E r .
The obvious question is whether we can calculate the derivative map
(G o F)* in terms of the derivatives G* and F*. The answer is yes. We state
the result, which should be intuitively obvious, and investigate some of its
consequences.
Theorem
For mappings F : E n ► E m and G : E m ► E r
(G oF), = G* o F*.
To be more precise we need to take into consideration the point at which each of
the derivatives is evaluated.
Now
F* p :T p (En) — T F(p) (Em)
G*F(p) : T F (p)(E m )— ^T G oF(p)(E r )
(G o F)^ : T p (En) — . T G F(p) (E')
and hence (Go F) 5ep = G^p) o F^ p . (This may remind you of the composite rule
for ordinary differentiation — D(g p f) = (Dg ° f).Df .)
Considering all points p of the domain at the same time we can write the composite
rule as
(GoF) Jle = G* F oF*.
This form of the composite rule accounts for all of the composite rules that we have
encountered. When we expand it in terms of Jacobian matrices the composition on
the right hand side then stands for the usual matrix multiplication.
M334I.7
47
(i) Given maps f : R ► R and g : R ► R, the cgmposite rule gives the form
of the chain rule encountered in Ml 00.
(ii) Given a map h : R ► R followed by a : R ► E 3 , as in Lemma 1.4.5, the
composite rule gives
(iii)
(a°h)*s = a*h(s) oh * s 
Now h, = and a, is represented by
ds
dt
AOLr
dt
doc 3
\dt
Hence if we write ft = a o h the composite rule gives
/*(.)\ . /^(h(s))\ .dh (s)
/ dt /dt I ds
d£
dt
Ms)
(Ms))
dt
dt
</ I
*(.)/ \*S(h(.))
dt
dh
i.e.0'(s)=.^i(s)a'(s).
ds
Given a map f : E 3 ► R followed by h : R
write the composite rule as
(h°f)* = h* f of,.
•+R, as in Lemma 1.5.7, we can
Now h* is just the derivative which can be written h', and so h* f = h'(f).
Hence in terms of Jacobian matrices
[ 9 (Mf)) j 3(h(f)) ] 3(h(f)) \ = h , (f) / 3f^ £_ af\
» 3x i 9x 2 ^x 3 / ^a Xl a X2 ' a x l'
i.e. WL) = h'(f)lL 0=1,2,3).
3xj dxj
(iv) Given a map a : R ► E 3 followed by a map f : E 3
then the composite rule states
■* R, as in Lemma 1.4.6,
( foO 0*t = f *a(t) oOf *f
48
M334 1.7
In terms of Jacobianjnatrices, this becomes
?m ( t) =(!L ( «(t)), 5. («(,)), ^_ Wt ))) /i«L(t)\
dt iBx! 8x 2 3x 3 // dt
so
dt i=l 3xj dt
Now read the proof of Lemma 1.4.6 and Lemma 1.3.2, where
a : 1 1 — ► p + tv, so that (dcq/dt)(0) =(d/dt)(pj + tv;)  t = q vj.
Summary
Notation
F = (fi,...,f m )
F*
Ibi df\
8u 8v
i 9 g ^g
\3u a7/
Definitions
(i) Euclidean coordinate functions (f l5 .
(ii) Image of curve under a mapping F(a)
(iii) The derivative map F*
(iv) Jacobian matrix
Results
(i) The
t derivative map can be describe
.fm)
the coordinate functions and directional derivatives
F*(v) = (v[f,],...,v[f m ]).
(ii) The derivative map F^p : T p (E n ) >T F (p) (E m )
is a linear transformation.
(iii) The effect of F* on the standard basis is
F*(Uj) = 2^L tV
i 9x;
Page 33, Definition 7.1
Page 35, Definition 7.4
Page 37 and Text, page 44
Page 33, Definition 7.1
Page 33, Definition 7.2
Page 35, Definition 7.4
Page 37 and Text, page 44
Page 36, Theorem 7.5
Page 36, Corollary 7.6
Page 37, Corollary 7.7
M334 1.7 49
(iv) The derivative map preserves velocities.
If = F(a) then 0' = F # (a'). Page 38, Theorem 7.8
(v) The composite rule:
If F : E n >E m and G : E m >E r then
(G° F )*p = G*F(p)°V
Techniques
(i) Description of mappings in terms of Euclidean co
ordinate functions. Page 33, Definition 7.1
(ii) Pictorial description of simple mappings.
(iii) Determination of derivative maps,
(a) from first principles Page 35, Definition 7.4
(b) from directional derivatives of coordinate
functions Page 36, Theorem 7.5
(c) from Jacobian matrix. Text, page 44
(iv) Calculation of the derivative mapping of a
composite using the composite rule:
(G°F)* = G* F °F*
where if the derivatives are represented by Jacobian
matrices the composition on the right is matrix
multiplication.
Exercises
Technique (i)
1. Express the following mappings from E 3 to E 3 in terms of Euclidean
coordinate functions
(a) F:p.— ►3p,
(b) F: P h— ► (ePiP2,p3+2 Pl ,pJ).
Technique (ii)
2. Page 39, Exercise 2.
Technique (iii)
3. (a) Page 39, Exercise 4.
(b) Find F*(v p ), using directional derivatives, for F = (x cosy, x siny, z),
v = (2,l, 3)andp = (0, 0, 0).
(c) Describe the derivative mappings for F given in Example (1), page 34.
50
Technique (iv)
M334 1.7
4. Use the composite rule for Jacobian matrices to find (9f/9x, 9f/9y, 9f/9z)
for the composite mapping in Exercise 4(a), page 6.
Solutions
(a) Since p • — ►  3p,
F(p) = (3pi, 3p 2 , 3p 3 )
= (3x(p), 3y(p), 3z(p))
and so, in terms of Euclidean coordinate functions,
F = (3x, 3y, 3z).
(b) F(p) = (ePiP2, p 3 +2p lf p\)
= ( e x(p)y(p), z(p) + 2x(p), (x(p)) 2 )
= (exy( p ),(z + 2x)(p),x 2 (p))
and so F = (e x Y, z + 2x, x 2 ).
Page 39, Exercise 2.
The lines u = 1, and v = 1 are the sets of points p such that u(p) = Pi = 1 and
v(p) = p 2 =1. These are shown on the following diagram.
V
u= 1
(1.1)
v= 1
(0,0)
►
A point on the horizontal line is of the form (t, 1) and is mapped by F to
(t 2  1, 2t). So, for instance, (0, 1) i — >(l, 0), (1, 1) " — >(0, 2) and
(1, 1)h>(0,2).
M334 1.7
51
(1,0)
► u
How do we describe such a curve? A point p= (pi,p2) is on the curve if
p t = t 2  1, p 2 = 2t for some t. From these identities we deduce that
4(pj + 1) = p 2 . Using the notation of Section 1.4 we write
pGM: 4(u+lj v 2 =0.
The line u = 1 consists of points of the form (1, t). These are mapped by
F to (1  t 2 , 2t). This is a typical point on the parabola v 2 = 4(u  1), which
is shown on the following diagram.
► u
3. (a) Page 39, Exercise 4(a).
F*(v p )=_(F( P + tv)) t = atF( P )
dt
= ( F (Pl + tV!, p 2 +tV 2 , p 3 +tV 3 )) t =
dt '
= — (Pi +tv i "P2 tv 2 ,p! +tvj +p 2 +tv 2 ,2p 3 +2tv 3 )l t = o
dt '
= (vi v 2 , Vl +v 2 , 2v 3 ) at F(p).
52
M334 1.7
Page 39, Exercise 4(b).
If F is linear then F(p + tv) = F(p) + tF(v).
Hen <* F*( v ) = l(F(p + tv))  t =0> ^ F(p),
dt
=l(F(p) + tF(v)) t=0 ,atF(p),
dt
= F(v) at F(p),
so F^Cvp) = F(v) F ( p ).
(b) If F = (x cos y, x sin y, z ), then by Theorem 1.7.5
F*(v p ) = (v p [xcosy], v p [xsiny], v p [z]) F ( p ).
Now if v = (2,1,3) and p = (0,0,0), then by Lemma 1.3.2
/\ rl B
v p [xcosy] = 2 (xcosy)(p)  — (xcosy)(p)+ 3 — (xcosy)(p)
dx 9y 3z
= 2 cosy (0,0,0) + x sin y (0,0,0) + 3.0(0,0,0)
= 2cos0 + OsinO + 3.0
= 2.
Similarly,
v p [xsiny] = 2 sin y(0,0,0)  xcosy(0,0,0) +
= 2sin0  OcosO = 0.
v p [z] = 2,0  1.0 + 3.1
* 3.
So
F*(v p ) = (2,0,0)( 0cos0>0 sin0, 0) = (2,0,3) (0,0,0)
(c) If F = (x  y, x + y, z) the Jacobian matrix for F^ is
1
Hi
dfi
Mi
3x
a y
dz
3f 2
3f 2
8f 2
3x
3y"
3z
df 3
3f 3
M3
8x
8y
3z
This maps the vector v to
/ 1 1
1
\
' V1 \
vi v 2 \
v 2 =
Vl +v 2
\ v 3 /
\ v 3
andsoF*(v p ) = (v t v 2 ,V! + v 2 , v 3 ) F ( p ), as expected.
M334 1.7
53
4. We have f described as a composite of functions, f = h o g, where
g : p i ► (gi(p), g 2 (p), g3(p)) That is, g u g 2 , g 3 are the coordinate functions
for g and so in this example
g = (x + y, y 2 , x + z).
Since f : E 3 ► R, g : E 3 ► E 3 and h : E 3 ► R, the corresponding
I : iL" ► K,
g:IV
►
m matnces are
(3f 9f
9f\
r
Ox 9y
9z/
dgi 9gi
3gi\
dx 9y
9z
a g2 3g 2
9g2
9x 9y
9z
2y
a g3 dga 9g 3
9x 9y 9z
, /9h 9h 9h\ /0
and [_, — , _q = (2x, z, y).
; 9x 9y 9zJ
Now by the composite rule f* = h* / \ © g^
so
i£(p), i£(p),H<p)U( 2x (g(p)), z(g( P )) ( y(g( P )))
9x 9v 9z / » /
1
2y(p)
1
(2gi(p),g3(p),g 2 (p)) l l 1 °
2p 2
1 1
(2pi + 2p 2 , Pl p3,p 2 2 ) /l 1
2p 2
1 1
= (2 Pl + 2p 2 pJ,2 Pl + 2p 2 2 Pl p 2 2p 2 p 3 , pj),
and hence
/9f 9f 9f\ /0
— > , — = (2x + 2y  y 2 , 2x + 2y  2xy  2yz, y 2 ).
\3x 9y 9z/
54 M334 1
FURTHER EXERCISES AND SOLUTIONS
Section 1. 1
Technique Exercises
Pages 56, Exercises 1(b), 2(a), 2(d), 4(b), 4(c).
Solutions
1(b). xy(2y + x)sin z
2(a).
2(d). t 4 (lt 3 )
4(b). — = 2e2x(i.
3x
eV)
4(c). ii = 4x.
3x
Section 1.2
Technique Exercise
Page 11, Exercise 4.
Other Recommended Exercise
Page 11, Exercise 5. This extends the concepts of linear independence and linear
combinations to cover vector fields.
Solutions
4. f = kx 2 , g = ky 2 for any real function k on R 3 ,
e.g. f = x 2 , g = y 2 .
5(a). V 1 (p) = (l,0,p 1 ) p
V 2 (p) = (0,l,0) p
V 3 (p) = (Pi,0,l) p
and since
1
Pi
1
Pi
1
1 + p? =£ these vector fields are linearly independent.
5(b). (xU! + yU 2 + zU 3 )(p) = (pi,p 2 ,p 3 )p
. PiUzPs) Vl (P) + P2 v 2 ( P ) + (pLLElV, (p)
(1+Pl) (1 + Pl)
and hence
xu t + y u 2 +zu 3 = x(1 ~ z) Vi +yv 2 + (x +z) v 3 .
(1+x 2 ) (1+x 2 )
M334 1 55
Section 1.3
Technique Exercises
Pages 1415, Exercises 1(a), (b), 2(a), (b), 3(a)(d), (f).
Solutions
1(a) and 2(a).
1(b) and 2(b). 896
3(a). y 3
3(b). 3xz 2
3(c). yz 2 (y 2 z3x 2 )
3(d). yz 2 (y 2 z3x 2 )
3(f).
Section 1.4
Technique Exercise
Page 21, Exercise 8.
Other Recommended Exercises
Page 21, Exercise 4. This exercise shows that a curve is determined uniquely by its
initial point and velocity vectors.
Page 21, Exercise 6. This theory exercise deals with the definition of the directional
derivative.
Page 21, Exercise 9. This deals with the geometrical interpretation of tangent lines.
Solutions
8. /3(s) = s,2,V21ogs),
0'(s) = (l,±,V£\
a'(h(s)) = /s,i,V2J.
dh, x 1 au
— (s) =_and hence j3'(s) = ™( s ).a'(h(s)).
ds s ds
4. a(t) = (l + l\il,et6).
3 2
56 M334 1
6. Let a be any curve with ol (0) = v p .
Thenfi!^l(0) = a(0)[f] = v p [f].
dt
9. (a) u i — ► (2, 2u, u).
(b) u h» (V2(l  u), V2(l + u),JL+ u).
Section 1.5
Technique Exercises
Pages 2526, Exercises 1(a), (b), 3(a), (c), 4, 6(a), (c).
Other Recommended Exercises
Page 25, Exercise 2. This expresses the operation of a 1form on a vector field in
terms of Euclidean coordinate functions.
Page 26, Exercise 8. This deals with the Leibnizian property of d.
Page 26, Exercises 9, 10. These explore the relationship between the differential of a
function and its maxima and minima.
Page 26, Exercise 11. This deals with the relationship between the differential and
the difference operator.
Solutions
1(a). 4
1(b). 4
3(a). x 3 +x 2 (l +z)xy 2 LI
x
4(a). 5f 4 df
df
4(b).
4(c).
2>/f
2fdf
1+f 2
6(a). df = y 2 dx + (2xy  z 2 )dy 2yz dz, 10.
6(c). df = cos(xy)cos(xz)(y dx + x dy)  sin(xy)sin(xz)(z dx + x dz), 2.
0(V) = 22 f iV ; dxi(U;) = 22 fjv^i; = 2 f iVi .
i j J J i j J J i
8 d(fg).(v p ) = v p [fg] = v p [f] g + fv p [g]
=df(v p )g + fdg(v p )
= (gdf + fdg)( Vp )
2
M334 1 57
and hence d(fg) = g df + f dg.
9. df(v p ) =* (pK +.21 (p)v 2 +iL(p)v 3
3x 3y 3z
and hence df(v p ) = for all v if and only if
i£(p)i!(p)>i£(p)o.
3x 3y 3z
Here Jl = 2xy.il = 1  x 2  2yz,J! = 1  y 2
3x 3y 3 Z
and the critical points are ± (0, 1, 5).
10. If p is a local maximum or minimum then t = is a local maximum or
minimum of the function 1 1 — ► f(p + tv).
Hence df [v p ] = v p [f] =_1 (f(p + tv))  t=Q = .
dt
11(a). By Taylor's Theorem
f(p + tv) = f(p) + t£(f(p + tv))  t=0 + Remainder
dt
= f(p)+tdf(v p )+R.
Hence f(p + v)  f(p) ~ df(v p ).
11(b). Exact: 0420
Approximate: df(v p ) = 1*?* dx +JL dy  £l dz) (01, 01, 0.2) (1> 1 . 5> x)
2
Section 1.6
Technique Exercises
Page 31, Exercises 1, 2, 4(a), (b), (d), 6.
Other Recommended Exercises
fhTd.?' ExerC r/; ^ "P*"" th e wedge product of three 1forms in terms of
the determinant of their Euclidean coordinates.
Page 31, Exercise 8. This exercise deals with the relationship between the exterior
derivative and the vector operations div, grad and curl.
Solutions
1(a). yz cosz dx dy  sinz dx dz  cosz dy dz,
sinz dx dy + z sinz dx dz + z cosz dy dz,
yz dx dy  yz 2 dx dz + dy dz.
58 M334 1
1(b). z dx dy  y dx dz,
cos z dx dz + sinz dy dz,
0.
2. d0 = dx dy, d\}/ =  dy dz,
y 2
0a \p = dx dy and
y
d(0Ai//) = d0A^  0Adi/ = dx dy dz.
y
4(a). d(fdg + gdf) = d(d(fg)) = 0.
4(b). (df  dg)A(df + dg) = 2 dfAdg.
4(d). (lf)dfAdg
6. dx dy dz = r dr d& dz
5. Expand both the determinant and the triple wedge product in terms of the
Euclidean coordinate functions. Both sides give
(f 11^33 ~ fllf 2 3f32 + ^23131 " ^2^33 + f 1^2^32 " * 13^22^1 )dx dy dz.
8(a). df = zfl dxi t W, 2^1 Ui = grad f.
9x • 3xj
8(b). If V = Sfi Ui then = Sfj dxj,
d =(^ !M d XldX2 + (2f! !M d Xl dx 3 + [Hl ^)dx 2 dx 3
and d0 < ( 2 ) , curl V.
8(c). If V = 2fiUjthenT7 = f 3 dx 1 dx 2  f 2 dxjdx 3 + f!dx 2 dx 3 .
dr?=Fi +^!i +^1\ dx, dx 2 dx 3 and hence dr? ( ( 2 ) , (div V)dx dy dz.
\dxi 9x 2 3x 3 ;
Section 1.7
Technique Exercise
Page 39, Exercise 1.
Other Recommended Exercises
Page 40, Exercises 5 and 6. These describe the derivative of a mapping in terms of
the differentials of its Euclidean coordinate functions.
Page 40, Exercise 8. This deals with the definition of the derivative.
Page 40, Exercise 9. The derivative mapping preserves directional derivatives.
M334 1 59
Page 40, Exercise 11. This deals with inverse mappings, which you can read about in
Exercise 10.
Page 40, Exercises 7 and 12. These deal with the derivative of a composite mapping.
Solutions
1(a). (0, 0).
1(b). (3,1), (3,1).
1(c). (0,0), (1,0).
5. F^ = (cos y dx  x sin y dy, sin y dx + x cos y dy, dz)
(a) (2,0,3) (0) o,0)
(b) (2,2,3) (0 ,2,7T)
6. The Jacobian matrix at p is
/ cos p 2 sin p 2
p 1 siwp 2 piCosp 2
,0 1
This has determinant pj and hence F is not regular at points for which
Pi=0.
Let a be any curve such that a (0) = v p , then if )3 = F(a) it follows that
j8 , (0) = F*(a , (0)) = F«(v p ).
Let a : 1 1 ► p + tq.
R^E n iE m _l,R
d
v p [g(F)] = a'(0)[g(F)] = ^(g(F( a ))) t=0 .
dt
F*(v p ) = (F(a))'(0).
F*(v p )[g] =(F(*))'(0)[g] =l(g(F(<*))) t=0 .
dt
Hence v p [g(F)] =F*(v p )[g].
11(a). F 1 = (v,ue" v ). This is a diffeomorphism.
L JL
11(b). F" 1 = (u , v + u ). This is not a diffeomorphism since it is not differentiable
when u = 0.
11(c). F" 1 = ( ( " u ~ 2v ) , 5  u  v. This is a diffeomorphism.
7 GF = (g 1 (f 1 ,f 2 ), g2 (f 1 ,f 2 )).
12(a). GF = (g 1 (f 1 ,...,f m ),...,g p (f 1> ..., fm))
60 M334 1
12(b). (GF)*(cO =? ((GF)(a))\ by Theorem 1.7.8,
=(G(F(*)))'
=G*((F(c*))')
=G*F*(c*').
12(c). Since any v p = a'(0) for some a it follows that (GF)* = G*F*.
M334 I 61
DIFFERENTIAL GEOMETRY
I Calculus on Euclidean Space
II Frame Fields
III Euclidean Geometry
IV Calculus on a Surface
V Shape Operators
VI Geometry of Surfaces in E 3
1§< PA^TII THE OPEN UNIVERSITY
!
I Mathematics: A Third Level Course
9
DIP
PART II FRAME FIELDS
DIFFERENTIAL GEOMETRY
9
THE OPEN UNIVERSITY
Mathematics: A Third Level Course
DIFFERENTIAL GEOMETRY PART II
FRAME FIELDS
O.U. COURSE U NITS COLLECTION
FOR REFERE^'cToNLY
A commentary on Chapter II of O'Neill's
Elementary Differential Geometry
Prepared by the Course Team
THE OPEN UNIVERSITY PRESS
Course Team
Chairman:
Mr. P.E.D. Strain
Dr. R.A. Bailey
Lecturer in Mathematics
Course Assistant in Mathematics
With assistance from:
Dr. J.M.
Mr. G.J.
Dr. P.M.
Mr. P.B.
Dr. F.C.
Mr. T.C.
Mr. R.J.
Dr. C.A.
Mr. M.G
Aldous
Burt
Clark
Cox
Holroyd
Lister
Margolis
Rowley
. Simpson
Senior Lecturer in Mathematics
Lecturer in Educational Technology
Lecturer in Physics
Student Computing Service
Lecturer in Mathematics
Staff Tutor in Mathematics
Staff Tutor in Mathematics
Course Assistant in Mathematics
Course Assistant in Mathematics
Consultants:
Prof. S. Robertson
Prof. T. Willmore
Professor of Pure Mathematics,
University of Southampton
Professor of Pure Mathematics,
University of Durham
4AH CXMM. ^o,
4
The Open University Press, Walton Hall, Milton Keynes
First published 1975
Copyright © 1975 The Open University
All rights reserved. No part of this work may be reproduced in any form, by mimeograph or any other means
without permission in writing from the publishers.
Produced in Great Britain by
The Open University Press
ISBN 335 05700 4
This text forms part of the correspondence element of an Open University Third Level Course. The complete
list of parts in the course is given at the end of this text.
For general availability of supporting material referred to in this text, please write to the Director of
Marketing, The Open University, P.O. Box 81, Milton Keynes, MK7 6AT.
Further information on Open University courses may be obtained from The Admissions Office, The Open
University, P.O. Box 48, Milton Keynes, MK7 6AB.
1.1
M334TI
CONTENTS Page
Set Book 4
Bibliography 4
Conventions 4
II. 1 Introduction and Dot Product 5
II.2 Curves 16
II. 3 The Frenet Formulas 28
II.4 ArbitrarySpeed Curves 38
II. 5 Covariant Derivatives 49
II. 6 Frame Fields 55
II. 7 Connection Forms 59
II. 8 The Structural Equations 61
Partial Differentiation 66
FurtherXxercises and Solutions 72
4 M334 II
Set Book
Barrett O'Neill: Elementary Differential Geometry (Academic Press, 1966). It is
essential to have this book: the course is based on it and will not make sense without
it.
Bibliography
The set books for M201, M231 and MST 282 are referred to occasionally; they are
useful but not essential. They are:
t).L. Kreider, R.G. Kuller, D.R. Ostberg and F.W. Perkins: An Introduction to
Linear Analysis (AddisonWesley, 1966).
E.D. Nering: Linear Algebra and Matrix Theory (John Wiley, 1970).
M. Spivak: Calculus, paperback edition, (W.A. Benjamin/AddisonWesley 1973).
R.C. Smith and f. Smith: Mechanics, SI edition (John Wiley, 1972).
Also mentioned in this Part is:
T. Wilmore: An Introduction To Differential Geometry (O.U.P., 1964).
Conventions
Before starting work on this text, please read M334 Part Zero. Consult the Errata
List and the Stop Press and make any necessary alterations for this chapter in the set
book.
Unreferenced pages and sections denote the set book. Otherwise
O'Neill denotes the set book;
Text denotes the correspondence text;
KKOP denotes An Introduction to Linear Analysis by D.L. Kreider, R.G.
Kuller, D.R. Ostberg and F.W. Perkins;
Nering denotes Linear Algebra and Matrix Theory by E.D. Nering;
Spivak denotes Calculus by M. Spivak;
Smith denotes Mechanics by R.C. Smith and P. Smith.
References to Open University Courses in Mathematics take the form:
Unit Ml 00 22, Linear Algebra I
Unit MST 281 10, Taylor Approximation
Unit M201 16, Euclidean Spaces I: Inner Products
Unit M231 2, Functions and Graphs
Unit MST 282 1, Some Basic Tools.
M334 II. 1
II. 1 INTRODUCTION AND DOT PRODUCT
Introduction
This whole section depends only on Sections 1.1 and 1.2, and may be read easily
even if you are not entirely confident about the later sections of Chapter I.
It provides a revision of the basic vector concepts that you have met in MST 282 and
M201. You should make sure that you are familiar with the following ideas: the dot
product, cross product, triple scalar product and norm for E 3 (Unit MST 282 1,
Some Basic Tools, Section 2); the dot product, norm, Schwarz inequality, angle,
orthogonality, orthonormal bases and orthonormal expansions for general Euclidean
vector spaces (Unit M201 16, Euclidean Spaces I: Inner Products); orthogonal
matrices (Unit M201 24, Orthogonal and Symmetric Transformations, Section 1.2).
You will also find it useful to revise the technique of evaluation of a 3 X 3 deter
minant (KKOP, page 684).
These vector ideas are revised and then applied to the tangent vector spaces Tp(E 3 ).
For example, the dot product of two tangent vectors with the same point of appli
cation is defined by
v w = v*w.
v p w p
We find that the usual vector space results carry over to each of the tangent spaces.
Later on we shall extend these definitions to vector fields by the pointwise
principle: for example, if V and W are vector fields the dot product of V and W,
V'W, will be the function from E 3 to R whose value at p is V(p)'W(p).
READ: Introduction to Chapter II and Section II. 1 (pages 4248).
Comment
(i) Page 47: the determinant and cross product The determinant is a
function from the set of all square real matrices into the real numbers. It is
written by replacing the brackets of the matrix by straight lines. The rules
for evaluating it on 2 X 2 and 3X3 matrices are as follows:
a ll a 12
a 21 a 22
a ll a 22  a 21 a 12
a ll
a 21
■■12
L 13
a 22 a 23
a 31 a 32 a 33
= a ll a 22 a 33 +a 21 a 32 a 13 +a 31 a 12 a 23
 a 31 a 2 2a 13  a2ia 12 a 3 3  a n a 32 a 23
(see KKOP, page 684)
which may be written as
a n a i2
L 13
a 21 a 22 a 23
a 31 a 32 a 33
= a
11
a 22 a 23
a 32 a 33
+ a
12
+ a
13
M334 II. 1
a 23 a 21
a 33 a 31
a 21 a 22
a 31 a 32
v 2
v 3
+ U 2 (p)
v 3
Vl
+ u 3 ( P )
Vl
V 2
w 2
w 3
w 3
w 2
Wj
w 2
This last expression gives the expansion of the formal determinant for v X w
as
U!( P )
= (v 2 w 3 v 3 w 2 ) U!(p) + (v 3 wx  v 2 w 3 ) U 2 (p) + (v,w 2 V 2 W!) U 3 (p)
= (v 2 w 3  v 3 w 2 ,v 3 w 1 v 1 w 3 , Vl w 2  V 2 W!) p .
O'Neill assumes that you are familiar with the following properties of deter
minants:
(a) a determinant* is linear as a function of each of its rows;
(b) interchanging two rows of a determinant multiplies its values by  1;
(c) a determinant has value zero if and only if its rows are linearly
dependent vectors.
In Unit M 201 5, Determinants and Eigenvalues, you learnt these properties
for the columns of a determinant. However, in Section 1.5 of that unit you
saw that the determinant of the transpose of a matrix is the same as the
determinant of the original matrix, and thus all these results are true for
rows as well.
By Lemma II. 1.8, v X w is orthogonal to both v and w, and
Hv X w  2 = (vv) (ww)  (vw) 2
= Hv 2 w 2 (vw) 2 .
The angle # between v and w is defined by
v*w= vw COS #
(see page 44), and so
llv X w 2 = v 2 w 2  (v Hwll cos I?) 2
= llv 2 w 2 (lcos 2 #)
= llv 2 w 2 sin 2 #.
M334 III
Taking square roots,
llvXwII =vw(sin 2 0) 2 .
1
.2 A x2
For any given value of cos # (in [ 1,1] ) we can choose # to lie in the range
< # < 7T, that is # is the smaller angle between v and w; for # in this range
sin # is nonnegative and so (sin 2 dp = sin &. Thus
Iv X wll = llvllllwll sin #
and so
v X w = vw sin # n
where n is a unit vector orthogonal to both v and w.
v X w
► w
In fact n is obtained from v and w by the righthand rule (page 48, penulti
mate paragraph), and so O'Neill's definition of crossproduct is equivalent to
that given in Unit MST 282 1, Some Basic Tools.
One consequence of the definition of crossproduct is the following.
Lemma II. 1. A If c 1 and e 2 are orthogonal unit tangent vectors and
e 3 = e x X e 2 then e 3 is a unit tangent vector orthogonal to both e! and e 2
and
e i X e 2 = e 3 ;
e 2 X e 3 = ej ;
e 3 X ei = e 2 ;
e 2 X e l = e 3 ;
e 3 X e 2 = e x ;
e i X e 3 = e 2 .
In particular, these formulas hold when e x =U 1 (p), e 2 = U 2 (p), in which
case e 3 = U x (p) X U 2 (p) = U 3 (p).
8 M334 II. 1
There is no reason to restrict the definition of cross product to tangent
vectors. As for the dot product, we can define it for pairs of points of E 3
also. If we put U! = (1, 0, 0), u 2 = (0, 1, 0), u 3 = (0, 0, 1) then for points p
and q of E 3 we define the cross product of p and q to be
u 1 u 2 u 3
pX q= pi p 2 p 3
qi q2 q3
= (P2q3 P3q2)«i +(p3qi piqsK + (piq2 P2qi)u 3
= (P2qs Psqa.psqi Piq3>piq2 P2qi)
Useful results about determinants and cross products are contained in the
following exercises, which you should attempt now.
1. Page 49, Exercise 4.
2. Page 49, Exercise 5 (first part).
3. Page 49, Exercise 6.
Additional Text
The result of the following exercise will be needed later.
4. Page 49, Exercise 7.
Summary
Notation
pq
vpw p
IIpII
llvpll
d (p> q)
*A
v p X w p
pXq
Page 42, Definition 1.1
Page 44, Definition 1.3
Page 43, line 9
Page 44, line 12
Page 43, Definition 1.2
Page 43, line 4
Page 47, line 1
Page 47, Definition 1.7
Text, page 8
Definitions
(i) Dot product of points p*q
of tangent vectors vw
(ii) Norm of a point p
of a tangent vector v
(iii) Euclidean distance d(p, q)
Page 42, Definition 1.1
Page 44, Definition 1.3
Page 43, line 9
Page 44, line 12
Page 44, Definition 1.2
M334 II. 1
(iv) e neighbourhood 9l (
(v) Open set
(vi) Orthogonal vectors
(vii) Unit vector
(viii) Frame, or orthonormal basis
(ix) Orthonormal expansion
(x) Attitude matrix
(xi) Orthogonal matrix
(xii) Transpose of A, t A
(xiii) Cross product of points pXq
of tangent vectors v X w
(xiv) Triple scalar product' u«v X w
Page 43, line 4
Page 43, line 3
Page 44, line 7
Page 44, line 6
Page 44, Definition 1.4
Page 45, line 1
Page 46, Definition 1.6
Page 46, line 2
Page 47, line 1
Text, page 8
Page 47, Definition 1.7
Page 48, line  3 of text
Results
(i)
lip + qll < llpll + llqll (the triangle inequality) and
llapH = a llpll.
(ii) vw < v w (Schwarz inequality).
(iii) Uj(p), U 2 (p), U 3 (p) constitute a frame at p.
(iv) Orthonormal expansion. If e^ e 2 , e 3 is a frame
at p and v€T p (E 3 ), then
v = (ve 1 )e 1 + (ve 2 )e 2 +(ve 3 )e 3 .
(v) If e lt e 2 , e 3 is a frame, v = 2 ajej and
w = Z bjej, then
vw = a^! + a 2 b 2 + a 3 b 3 .
(vi) If A is an orthogonal matrix, *A = A 1 .
(vii) The dot product is symmetric, bilinear, positive
definite.
(viii) The cross product is alternating, bilinear.
(ix) v X w = (vv ww (vw) 2 )2 = v w sin #.
(x) v X w is orthogonal to v and to w.
(xi) v X w = vj w sin # n, where n is obtained by the
righthand rule.
(xii) v X w # if and only if v and w are linearly inde
pendent.
(xiii) u«v X w ¥= if and only if u, v and w are linearly
independent.
(xiv) Interchanging any two of u, v, w reverses the sign
of uv X w but does not change its absolute value.
(xv) If ej, e 2 , e 3 is a frame,
e l e 2 X e 3 =± 1.
Page 43, line 11
Page 44, line 13
Page 45, line 6
Page 45, Theorem 1.5
Page 46, line 7
Page 47, line 5
Page 43, lines 17
Page 47, line 11
Page 47, Lemma 1.8 and
Page 48, line 9
Page 47, Lemma 1.8
Text, page 7
Page 49, Exercise 5
Page 49, Exercise 4(b)
Page 49, Exercise 4(c)
Page 49, Exercise 6
10
(xvi) If e 1} e 2 are orthogonal unit vectors and
e i X e 2 = e 3 , then ej, e 2 , e 3 is a frame, and
M334 II. 1
ej X e 2 = e 3 ; e 2 X ei = ~e 3 ;
e 2 X e 3 = ei; e 3 X e 2 = e 1 ;
e 3 ^ e i = e 2> Ci X e 3 =  e 2 .
Text, page 7
(xvii) If u is a unit vector, any vector v may be expressed
uniquely as
V = (V'U) u + v 2 ,
where v 2 is orthogonal to u. Page 49, Exercise 7
Techniques
(i) Evaluation of dot and cross products and norms.
(ii) Evaluation of triple scalar products.
(iii) Expansion in terms of an orthonormal basis.
(iv) Determination of the attitude matrix and use of
the result t A = A" 1 .
Page 42, Definition 1.1 and
Page 47, Definition 1.7
Page 49, Exercise 4(a)
Page 45, Theorem 1.5
Page 46, Definition 1.6
Exercises
Technique (i)
5. Page 48, Exercise 1.
Technique (ii)
6.
If v and w are as in the preceding exercise and u = (3, 1,1) at the same
point, compute u«v X w.
Technique (iii)
7. Page 49, Exercise 3.
Technique (iv)
8. Find the attitude matrix A of the frame in the preceding exercise. What is its
inverse?
M334 II. 1
Solutions
11
1. Page 49, Exercise 4.
(a) Suppose u, v, w are tangent vectors at the point p.
u = U!Ui(p) + u 2 U 2 (p) + u 3 U 3 (p).
v X w = cjU^p) + c 2 U 2 (p) + c 3 U 3 (p) (*)
where the cj are found from
v X w =
iMp) u 2 ( P ) u 3 ( P )
Vi v 2 v 3
Wj w 2 w 3
UV X W = CjUj + C2U2 + c 3 u 3
which is (*) with Uj(p) replaced by uj for i = 1, 2, 3: thus replacing
Uj(p) by u[ in the determinant gives u«v X w.
(b)
(c)
(d)
u«v X w # <=*• the rows of the determinant are linearly independent
<==>• u, v and w are linearly independent.
Again, this follows from the properties of determinants: if any two
rows in a determinant are interchanged the whole determinant
changes sign.
u X v«w = wu X v,
= u«v X w,
because the dot product is symmetric,
by part (c).
Page 49, Exercise 5.
We know that lv X w = v w sin #, where # is the angle between v and w.
v X w = <==> v X w =
v = or w = or sin # =
v = or w = or & is a multiple of it.
If v = or w = then v and w are certainly linearly dependent. If # is a
multiple of tt then
w
w
# =
# = IT
v and w are collinear, and so linearly dependent. Conversely, if v and w are
linearly dependent, either v = or w = or sin # = 0.
Thus v X w t£ <=> v and w are linearly independent.
12
3.
M334 II.l
Page 49, Exercise 6.
e 2 X e 3 is orthogonal to both e 2 and e 3 . e 2 and e 3 span a plane in T p (E ),
so the tangent vectors orthogonal to both e 2 and e 3 form a line in T p (E 3 ).
We know that e 2 is orthogonal to both e 2 and e 3 and so e x spans this line:
thus e 2 X e 3 must be a multiple of e x .
lle 2 Xe 3  = e 2 e 3 sin=l,
so
Thus
e 2 X e 3 = ±e v
eie 2 X e 3 = ie^ej =±1.
If A is an orthogonal matrix, then it is the attitude matrix of a frame e^
e 2 ,e 3 , where the coordinates of ej with respect to Ux(p), U 2 (p), U 3 (p) form
the ith row of A for i = 1, 2, 3. By the first part of the question,
e 1 e 2 X e 3 = ± 1. However, by Exercise 4(a) on page 49,
e r e 2 X e 3
en e 12 e 13
e 21 e 22 e 23
e 31 e 32 e 33
= det A
so the determinant of A is ± 1.
We could have reached this last result in another way, using the facts (see
Unit M201 5, Determinants and Eigenvalues, Section 1) that if A and B are
square matrices of the same size then
and
det (*A) = det (A)
det (AB) = det(A) det(B),
Now, if A is orthogonal, *AA  I
so
so
so
so
det (*AA) = det(I) = 1
det (*A) det(A) = 1
det (A) det(A) = 1
det(A) = ±l.
Page 49, Exercise 7.
Choose v x = (vu)u, v 2 =v  v 2 . Then certainly v = Vj + v 2 and v x is the
component of v in the u direction, by definition. Moreover, v  \ x is the
only possible choice for v 2 in order that \ = \ x + v 2 , and so this expression is
certainly unique.
M334 II. 1
13
v 2
It remains to prove just that v 1 v 2 = 0. Now
vjvj =x l (x vO
= (vu)u • (v  (vu)u)
= (vu)(uv  (v«u)uu)
= (vu)(u«v  v«u), because uu = 1,
= 0, because u»v = v«u.
Page 48, Exercise 1.
(a) vw = l.( 1) + 2.0 + ( 1).3 =  1 3 = 4.
(b) U^P) U 2 (p) U 3 (p)
v X w= 1 2 1
10 3
= (2.3(l).0,(l).(l) 1.3,1.0 2.(l))
= (6,2,2).
(c) llvl = (i a +2 a +(i) a )*=V6; pr^
IMi = ((i) 2 + o 2 + 3 2 )^=ViO; IR^ 1 ^^
(d)
v Xw = (6 2 +(2) 2 +2 2 )=V44 = 2VH.
Alternatively, we can use Lemma II. 1.8:
v X w 2 = (vv)(ww)  (vw) 2 = v 2 w 2  (vw) 2 = 6.10(4) 2 = 44.
(e) If # is the angle between v and w,
v»w = jv w COS#
so 4 =^6/10 cos #
4 4 2
so costfsygoa^g^g
1
1
3
= 3(2.3 (1).0) + 1((1) 2  1.3) 1(1.0 2.(l))
= 3.6 + 1. (2) 1.2 =22.
Alternatively, since v X w has already been calculated in Exercise 5, we could
go straight to
uvX w = (3, 1,1). (6,2,2)
= 3.6 + l.(2) + (l).2 = 22.
Ul
u 2
u 3
3
1
•v X w =
Vl
v 2
v 3
=
1
2
Wj
w 2
w 3
1
14
7.
M334.II.1
Page 49, Exercise 3.
e 1 .e 1 = (l 2 +2 2 +l 2 ) = l
e 2 e 2 =((2) 2 + Q 2 +2 2 ) = 1
e 3 .e 3 =!(l 2 +(l) 2 +l 2 ) = l
ei ' e2= 7678 (  2 + + 2) =
e 2*e 3 = 7575 (2 + + 2) =
1 1
(12 + 1) =
e3 ' ei ~VV6
This shows that e 1} e 2 , e 3 constitutes a frame.
By Theorem II. 1.5,
v = (v«c 1 )c 1 + (ve 2 )e 2 + (ve 3 )e 3 .
v. ei =L (6 + 2 i)= 7
ve
V6
(122) =
V6'
14
v ' e3= 73( 61  1)= 73 ;
14 4
'e, +— — e.
V V6 e i"V8 C2 V 3 " 3 *
We can check this by working out the righthand side directly: we obtain
7 (1,2,1) 14 (2,0,2) 4 (1,1,1)
V6 V6 7 s " V 8 \73" V 3
=^(1,2, l)^(2,0,2)+(l,l, 1)
~ C> Q' C + O '» ^» 9 + *' «' 3
= (6,l,l)=v.
8. We simply take the coordinates of c u e 2 , e 3 and write them as the rows of
A. Thus
1
V6 V 6 V 6
A =
2
V8
_J_ ^1 —L
V 3 V 3 V 3
M334 II. 1 15
Since A is an attitude matrix it is orthogonal, and so we know t A = A * : that
is, A~ * may be obtained from A by interchanging corresponding rows and
columns.
This gives
,i 
1
2
1
V6
V«
V3
2
1
V3
1
2
1
V6
V8
V3
16 M334 II.2
II.2 CURVES
Introduction
This section depends heavily on Sections 1.4 and II. 1 but not at all on the later
sections of Chapter I.
Here we take up the story of curves left in Section 1.4, and extend the definitions
and results of Section II. 1 to vector fields defined on curves. We immediately
introduce a way of differentiating these vector fields; in particular this gives us the
acceleration of a curve: the velocity and acceleration turn out to be the same as
those of a particle travelling along the curve, as given in Unit MST 282 2,
Kinematics, Section 1.1.
All of this is simply giving us enough technical apparatus to establish the Frenet
formulas, which are a core feature of the course, in the next section.
READ: Section II.2 (pages 5155) .
Comments
(i) Pages 5152: unitspeed reparametrization This particular type of
reparametrization is very important. From now on, whenever O'Neill mentions a
"unitspeed reparametrization" he means not just a reparametrization which
happens to have unit speed but a reparametrization of the type given in the proof of
Theorem 2.1. There is a useful symmetry in this situation. If j3 is a unitspeed
reparametrization of a we have
3(s) = a(t(s));
in this particular case s and t are inverse functions, and so we also have
j5(s(t))=a(t):
that is, a is also a reparametrization of ]3. We shall often make use of this fact.
Theorem 2.1 is important because it is often sufficient to know that a unitspeed
reparametrization of a curve ol exists, without computing it explicitly. It is this
theorem which allows many proofs to begin: "We may assume a has unit speed".
This considerably simplifies many things, but you should always check that the
remark is justified.
(ii) Page 52: Definition 2.2 A vector field on a curve should be distinguished
from a vector field on E 3 . It is true that if o: is a curve and V is a vector field on E 3
then the function Y defined by
Y(t)=V(a(t))
is a vector field on a (see the diagram opposite); however not every vector field on a
is necessarily the restriction of a vector field on E 3 . For example, it may happen
that a crosses itself, that is there are different values t x and t 2 such that
a(ti)=<*(t 2 ) = p;
M334 II.2
17
Y(t) = V(a(t))
if Y is a vector field on a it is not obliged to take the same values at t t and t 2 , but if
V is a vector field on E 3 then it has a unique value at p and so
V(a(t!)) = V(a(t 2 )) = V(p). A good example of this is provided by the
figureofeight curve given by
a(t) = (sint,^sin2t, 0) (tGR)
which crosses itself at = a(0) = a(ir): at this point the vector field a on a
does take different values, for a'(0) = (1, 1, 0) while a'{ir) = (1, 1, 0).
► x
(Hi) Page 54: Lemma 2.3 It is important at this point to be quite clear about
O'Neill's slightly unusual use of the words "constant", "straight line" and "parallel".
A curve a is a constant if it is of the form
a: ti ► c
that is, the image of a in E 3 is a single point. A curve a is straight line if and only if
it is of the form
a: 1 1 ► p + tq
for some points p and q with q ¥= 0. The image of a is certainly the line joining
p and p + q. Intuitively we might think of the curves
/3: 1 1 > p + 2tq
18 M334 II.2
and
7: 1 1 »p + t 3 q
as the same as a, because their images are also the straight line joining p and p + q,
but it is important to realize that, under Definition 1.4.1, they are different curves
from a. In fact j3 is the reparametrization of a by the function
h: t 1 — >2t
and 7 is the reparametrization of a by the function
k: t 1 ► t 3 .
Are j8 and 7 straight lines? Well j(3 is, because j(3 can be written
0= ti >p + t(2q),
but 7 cannot be written in the form
7: t 1 >a + tb
for any points a and b so 7 is not, technically speaking, a straight line, merely a
reparametrization of one.
Finally, "parallel" has the meaning given on page 6, that of having the same vector
parts. This is a more restrictive definition than that usually used in elementary
geometry. For example, if
V = (1, 0, 0) (0> 0> Q) y w
W = ( 1 '0.0)(1,1,0)
x =(2,0,0) (1>1>0)
• ►
y =(l,0,0) (1)1?0) v
then v is parallel to w but not x or y, even though they all point along the same
direction. For tangent vectors at the same point we say they are collinear if they
point along the same direction, that is if one is a scalar multiple of the other. The
vectors w, x and y are all collinear.
(iv) Curves — a Convention From now on we adopt the convention used by
Spivak that a function defined on a subset of E has the whole of E as its domain
unless otherwise specified.. Thus in Exercise 1 on page 55 the curve
cx(t) = (2t,t 2 ,£)
is taken to be the curve
a: t 1 >(2t, t 2 , y) (tGR).
M334 II.2 19
Additional Text
The following exercises introduce results which will be needed later.
Attempt them now.
1. Page 56, Exercise 7.
2. Y and Z are vector fields on a curve a. Prove that
(YX Z)' = Y'X Z + YX Z'.
Supplementary Comment
(i) Page 51: the "standard theorem of calculus" This is a combination of
several results from Spivak, which you should be able to accept.
Summary
Notation
v
s
Y
a"
Definitions
(i) Speed v
(ii) Arc length
(iii) Arclength function s
(iv) Arclength parametrization
(v) Unitspeed reparametrization
(vi) Orientationpreserving reparametrization
(vii) Orientationreversing reparametrization
(viii) Vector field (Y) on a curve
(ix) Euclidean coordinate functions of a vector field on
a curve
(x) Derivative of a vector field on a curve
(xi) Acceleration (a ") of a curve
(xii) Parallel vector field
(xiii) Collinear tangent vectors
Page 5 1 , line 4
Page 5 1 , line  7
Page 52, Definition 2.2
Page 54, line 7
Page 5 1 , line 4
Page 5 1 , line 1 1
Page 5 1 , line  7
Page 52, line 7
Text, page 16
Page 52, line 12
Page 52, line 11
Page 52, Definition 2.2
Page 53, line 9
Page 54, line 3
Page 54, line 7
Page 54, line  10
Text, page 18
20
Results
(i)
(iii)
(iv)
Any regular _curve has a unitspeed repara
metrization.
If (3 is a unitspeed reparametrization of a, then a is
a reparametrization of jS.
If a has unit speed, then s(t) = t.
The linearity and Leibnizian properties of
differentiation of vector fields on curves.
(aY + bZ)' = aY' +bZ'
(fY)' = f'Y + fY'
(YZ)' =Y'.Z + YZ'
(YXZ)' =Y'XZ + YX Z'.
(v) a is constant if and only if a = 0.
(vi) a is a straight line if and only if a =fc and a" = 0.
(vii) llY II is constant if and only if Y« Y' = 0.
(viii) Y is parallel if and only if Y' = 0.
(ix) Y(h)' = h'Y'(h).
M334 II.2
Page 51, Theorem 2.1
Text, page 16
Page 54, lines 1217
Text, page 19
Page 54, Lemma 2.3(1)
Page 55, Lemma 2.3(2)
Page 54, lines  14 to  12
Page 55, Lemma 2.3(3)
Page 56, Exercise 7
Techniques
(i) Calculation of arclength.
(ii) Finding an arclength function s = s(t).
(iii) Unitspeed reparametrization.
(iv) Expression of a vector field on a curve in terms of
its Euclidean coordinate functions.
(v) Algebra of vector fields on a curve.
(vi) Differentiation of vector fields on a curve
(a) from first principles
(b) using linearity and Leibnizian properties.
(vii) Calculation of velocity, speed, acceleration.
Page 5 1 , line 1 1
Page 51, proof of Theorem 2.1
Page 51, Theorem 2.1
Page 53, line 7
Page 53, last paragraph
Page 54, line 3
Page 54, lines 1217 and
Text, page 19
Page 5 1 , line 4 and
Page 54, line 7
Exercises
Technique (i)
3. Page 55, Exercise 4. (Note that "log" means "natural logarithm".)
Techniques (ii) and (iii)
4. Page 55, Exercise 3.
M334 II.2 21
Techniques (iv) and (v)
5. Page 55, Exercise 6.
Techniques (vi)(a) and (vii)
6. Page 55, Exercise 2. The required cone is the set of points in E 3 satisfying
x 2 + y 2 = z 2 .
Technique (vi)(b)
7. Let Y and Z be the vector fields on the curve a(t) = (cos t, sin t, t) given by
Y(t) = (cost, sint, l) a (t)'
Z(t) = (sin t, t, cos t)a( t ).
Find Y'(t) and Z'(t) and hence V(t) when V is each of the following:
(a) V(t) = Y(t) + Z(t)
(b) V(t) = t 2 Y(t)
(c) V = YZ
(d) V = Y X Z.
Theory Exercises (omit if you are short of time).
8. Y is a vector field on a curve a. Prove that if Y is constant and Y' is never
zero then Y" is never orthogonal to Y. (HINT: Differentiate Y"Y = con
stant.)
9. Page 56, Exercise 9. (HINT: For the "if" part of (b), let 7 be a unitspeed
reparametrization of a and show that 7 is a straight line by showing that 7
is both collinear and orthogonal to 7'. By result (ii) a is a reparametrization
of 7.)
Solutions
1. Page 56, Exercise 7.
Y(t) is a tangent vector at a(t), so Y(h(s)) is a tangent vector at a(h(s)), so
Y(h) is a vector field on the curve a(h).
We can write Y(t) = (yi(t), y 2 (t), y 3 (t)) a ( t )
so
Y , (t) = (y' 1 (t),yi(t),y , 3 (t)) a ( t ).
22 M334 II.2
Now
Y(h)ts) = ( yi (h(s)), y 2 (h(s)), y 3 (h(s))) a(h(s))
so
Y(h)'(s) = (y'i(h(s)) h'(s), yi(h(s))h'(s), yg(h(s))h'(s)) a(h(s))
by the definition of
differentiation of a
vector field on a curve
and by the chain rule
= h '( s )(yi( h (s)), yi(Ms)), y3(h(s))) a (h( s ))
= h'(s)Y'(h(s))
so
Y(h)' = h'Y'(h).
Note that this result is, not surprisingly, yet another form of the chain rule!
2. If
Y(t) = ( yi (t),y 2 (t),y 3 (t))
and
Z(t) = ( Zl (t), z 2 (t), z 3 (t))
then
(Y X Z) (t) = (y 2 (t)z 3 (t) y 3 (t)z 2 (t), y 3 (t) Zl (t)  Yl (t)z 3 (t),
yi(t)z 2 (t)y 2 (t)z 1 (t)).
To differentiate Y X Z we differentiate its coordinate functions. The first of
these is
y 2 z 3 y 3 z 2 ,
whose derivative is
y 2 z 3 + y 2 z 3  y 3 z 2  y 3 z 2 , by the Leibnizian rule for derivatives
of functions from R to R,
= (Y2Z3 ~ y'&l) + (Y2Z3 " Y3Z2)
The other two coordinate functions are similar, and so we obtain
(Y X Z)'(t) = (y 2 (t)z 3 (t)  y 3 (t)z 2 (t), y 3 (t) Zl (t)  yi(t)z 3 (t) ,
y'i(t)z 2 (t)y 2 (t) Zl (t))
+ (y 2 (t)z 3 (t)  y 3 (t)z 2 (t), y 3 (t)zi(t)  yi (t)z 3 (t),
yi(t)z 2 (t) y 2 (t)zl(t))
= Y'(t) X Z(t) + Y(t) X Z'(t)
= (Y'XZ + YX Z')(t).
Thus
(YXZ)' = Y' X Z + YX Z'.
M334 II.2 23
3. Page 55, Exercise 4.
a(t) = (2t,t 2 ,logt),t>0.
cx(l)=(2, l,0)=p
<x(2)=(4,4,log2) = q,
so the curve passes through p and q.
a'(t) = (2,2t,i) a(t)
la'(t) = (4 + 4t 2 + p )"* = 2t +i since t > 0.
The arclength between p and q is the arclength from t = 1 to t = 2, which is
"2
i
2t + i dt =
t 2 +logt
2
J 1
= 4 + log 2  1
3+ log 2.
4. Page 55, Exercise 3.
<x(t) = (cosh t, sinh t, t)
c/(t) = (sinh t, cosh t, 1) a /j\
a'(t) = (sinh 2 t + cosh 2 t + 1)3 = (2 cosh 2 t)i
= >/2 cosh t.
because 1 + sinh 2 t = cosh 2 t,
s(t) = \ a' (u)  du = I J 2 cosh u du = J 2 sinh t
Jo JO
so
t(s) = sinh * —t~.
The unitspeed reparametrization of a. based at t = is
0(s) = a(t(s)) = (cosh(sinh _1 ~tt), sinh(sinh _1 ~jz), sinh 1 j^)
(using the identity cosh 2 =1 + sinh 2 for 6 = sinh * ~J7r),
24 M334 II.2.
5. Page 55, Exercise 6.
(a) Y(t) =  a(t) at a(t)
= ( cos t,  sin t,  1) a u\
= cost U^aft))  sint U 2 (a(t))  t U 3 (a(t))
=  cos tU] sin t U 2  t U 3
using the convention mentioned in the middle of page 53.
(b) a'(t) = (sint, cost, l) a ( t )
a"(t) = (cos t, sin t, 0) a u\
Y(t) = (cost  sintJUj + (cost + sint)U 2 + U 3 .
(c) Any tangent vector at a(t) orthogonal to both a'(t) and a"(t) is a
scalar multiple of a'(t) X a '(t), if this is not zero.
a'(t) X <x"(t) = (sint, cost, 1) a u\
oc'(t) X a"(t) 2 = sin 2 t + cos 2 t +1 = 2
so a unit vector in the direction of a'(t) X a"(t) is ± (ljy/2)oi'(t) x a " ( l )
Thus Y(t) = ± ( sin L U '  cos / t U 2 + Ih.
W \ V2 V2 V2
(d) Y(t) = (a(t+ir)a(t)) a(t)
= ((cos(t + 7r), sin(t + 7r), t + 7r)  (cos t, sin t, t)) a ( t \
= ((cos t, sint, t + 7r)  (cos t, sin t, t)) a u\
=  2cos t Ui  2sin t U 2 + 7rU 3 .
Page 55, Exercise 2.
Let p = a(t) = (t cos t, t sin t, t).
The usual reason for the presence of the sine and cosine functions in the
Euclidean coordinate functions of a curve is to take advantage of the
identity sin 2 + cos 2 = 1. Thus we see that
p 2 + p 2 = t 2 (cos 2 1 + sin 2 1) = t 2 = p 2 ,
that is, every point p on the curve satisfies
Pi + P^=P 2 3  (*)
M334 II.2 25
If C is the set of points satisfying (*), C intersects the plane p 3 = r in the
circle p 2 + p 2 = r 2 , for each real number r. Thus C is the circular cone with
axis the zaxis, vertex the origin, and halfangle — . Every point of the curve
lies on C; the vertex (0, 0, 0) is a(0).
a'(t) = (cos t  t sin t, sin t + t cos t, 1) a ^ t )
a'(0) = (1,0,1) ( o,o,0)
Ha'(0) = (l 2 +l 2 f =V2;
a"(t) = ( 2sin t  t cos t, 2cos t  t sin t, 0) a / t \
a"(0) = (0,2,0) (0,0,0)
7. We first evaluate Y' and Z', differentiating each coordinate function
Y(t) = (cost, sint, l)o;(t)
Y'(t) = (sint, cost, 0) a n\
Z(t) = (sint, t, cost) a ( t \
Z (t) = (cost, 1, sint)^/^.
(a) V(t) = Y(t) + Z(t)
so V'(t) = Y'(t) + Z'(t), by linearity,
= (cos t  sin t, cos t + 1 ,  sin t) (x(t)
(b) V(t) = t 2 Y(t)
so
V'(t) = 2t Y(t) + t 2 Y'(t), by the Leibnizian property,
= 2t (cost, sint, l)(Wt) + t 2 (sint, cost, 0) a n\
= (2t cost t 2 sint, 2t sint + t 2 cost, 2t) a uy
(c) V(t) = Y(t)Z(t)
so
V'(t) = Y'(t)Z(t) + Y(t)Z'(t), by the Leibnizian property,
= ( sin t, cos t, 0) • (sin t, t, cos t) + (cos t, sin t, 1 ) • (cos t, 1 ,  sin t)
=  sin 2 1 + t cos t + + cos 2 1 + sin t  sin t
= cos 2 1  sin 2 1 + t cos t„
26 M334 il.2
(d) V(t) = Y(t) X Z(t)
so
V'(t) = Y'(t) X Z(t) + Y(t) X Z'(t), by the Leibnizian property,
= (sint, cost, 0) a / t \ X (sint, t, cos t) a u\
+ (cost, sint, l)a( t ) X (cpst, 1, sint) a ( t )
= (cos 2 1, sin t cos t,  1 sin t  sin t cos t) a n\ +
( 1  sin 2 1, cos t + sin t cos t, cos t  sin t cos t) a / t \
= (cos 2 1  sin 2 1  1 , cos t + 2sin t cos t, cos t  t sin t 2sin t cos t) a / t \
= ( 2sin 2 1, cos t + sin 2t, cos t  tsin t  sin 2t) a n\.
8. If Y has the constant value c, then
YY = c 2 .
Differentiation gives
Y'Y + YY' =
so
YY' = 0.
Further differentiation gives
Y'Y' + YY" = 0.
If Y(t) is orthogonal to Y"(t) then (Y'Y")(t) = 0. This forces (Y'"Y')(t) = 0,
that is  Y (t) = 0. The norm of a vector is zero if and only if that vector is
itself zero, so this implies Y'(t) = 0, which is contrary to assumption. Thus
Y(t) cannot be orthogonal to Y"(t).
Page 56, Exercise 9.
(a)  \a  is a constant
a a is a constant
(«'.<*')' =
a a + a. a =0
a a =0
a is always orthogonal to a .
M334 II.2
27
(b) Suppose j3 is the straight line t>> p + tq, and a is the reparametrization 0(h)
of j3. Then
a(u) = p + h(u)q
a'(u) =h'(u)q a ( u )
a"(u)=h"(u)q«( u ).
Then a'(u) and a"(u) are both scalar multiples of the tangent vector q a ( u ),
and so they are collinear. Conversely, we wish to show that the collinearity
of a and a" implies that a is a reparametrization of a straight line. By the
remark about symmetry in Comment (i), it is enough to show that a has a
unitspeed reparametrization 7 = a(h) which is a straight line. Consequently
we choose 7 to be a unitspeed reparametrization of a: Theorem II.2.1 tells
us that we can choose. such a 7.
Now 7(s)  a(h(s))
7'(s) = h'(s) a'(h(s)), by Lemma 1.4.5,
7"(s) = h"(s) a (h(s)) + (h'(s)) 2 a"(h(s)), by the Leibnizian property.
a" (h(s)) and a'(h(s)) collinear
=> a"(h(s)) is a scalar multiple of a'(h(s))
=> t"(s) is a scalar multiple of ot (h(s))
=> 7"(s) and 7'(s) are collinear.
However, 7 has unit speed, that is \\y' \\ = 1, so part (a) tells us that 7"(s) is
orthogonal to 7'(s). The only way that 7" (s) can be both collinear and
orthogonal to 7'(s) is that 7"(s) = 0: hence by Lemma 11.2.3(2) 7 is a straight
line.
28 M334 II.3
II. 3 THE FRENET FORMULAS
Introduction
This section carries straight on from Section II. 2. The Frenet formulas for unitspeed
curves are the first really important result in the course. You should try to become
thoroughly familiar with the concepts of tangent, principal normal, binormal,
curvature K and torsion r, which are collectively called the Frenet apparatus, and be
able to find them for any unitspeed curve.
We use these ideas to define the Frenet approximation to a curve, which is a type of
Taylor approximation. You have met Taylor approximations before in Unit Ml 00
14, Sequences and Limits II or Unit MST 281 10, Taylor Approximation, and Unit
M231 13, Taylor Approximations. The Frenet approximation is less important than
the Frenet apparatus, except in so far as it helps one to visualize the shape of a curve
near a point and suggests various characterizations of curves in terms of their
curvature and torsion. Such a characterization is a statement of the form:
j3 is a curve of a particular type (e.g. plane curve, cylindrical helix) if and
only if the curvature (k) and torsion (r) of j3 satisfy a particular condition
(e.g. r = 0; or t/k is constant).
The fact that there are such characterizations indicates why k and r are important
functions associated with a curve: we shall see just how important they are in
Chapter III. The section concludes with some characterizations of curves: you should
note the methods of proof as much as the results. Further characterizations follow
in the exercises and in the next section.
READ: Section II.3 (pages 5663) .
Comments
(i) Pages 578: curvature and principal normal The definition of K as T'
forces us to avoid points where k is zero, for at such a point not only is N undefined
but K may not be differentiable (because the squareroot function is not differenti
able at zero). Moreover, N may suddenly change from one side of the curve to the
other near a point where k is zero, as shown in the diagram opposite.
Consequently the footnote on page 57 is very important: we restrict our attention
to portions of curves where the curvature is positive.
It is possible to develop a theory where k is allowed to become zero' and even
negative, in which N is defined for many curves even at points where K = 0. In this
theory k and N are still differentiable at such points. O'Neill does not take this
approach (except for curves in E 2 — see the optional section in the exercises), but if
you are interested you can read about it in T. J. Willmore: An Introduction to
Differential Geometry (O.U.P. 1964), Chapter I.
(ii) Page 57: binormal We choose B to be T x N specifically to obtain a vector
field which is orthogonal to T and N at each point. Since T(s) and N(s) are
orthogonal unit vectors, Lemma II. 1. A tells us that B(s) is a unit vector orthogonal to
T(s) and N(s), and moreover the following equations hold:
TXN = B; NXT=B;
NXB = T; BXN = T;
BXT = N; TXB = N.
M334 II.3
29
Since T(s), N(s), B(s) form a frame at each point 3(s) we can extend the idea of
orthonormal expansion by the pointwise principle to obtain the result that any vector
field Y on has an orthonormal expansion of the form
Y = (YT)T + (YN)N + (Y'B)B.
(in) Page 59 onwards: the correspondence between points and tangent vectors
The identification of the tangent vector v p with the point v, for working which is
concerned just with the Euclidean coordinates, enables us to omit the subscripts to
tangent vectors. This convention considerably simplifies manipulation and proofs.
For example, O'Neill uses this convention throughout the proof of Corollary 3.5:
without this convention the last line of page 6 1 would read
^( s )^jS(s) = ns)q j g(s) =
and the second paragraph on page 62 would contain words to the effect that since B
is parallel there is a fixed vector v such that B(s) = v^/ s \ and would then define f(s)
as (j3(s)  j8(0))«v; this adds verbiage and symbols to the proof without enhancing our
understanding of it. Consequently we use this convention whenever it is possible to
do so without causing confusion. We note that by page 58 O'Neill is already writing
tv \ _ i a. • s a s b
T(s) 7r sm,cos, j
c' c
rather than the technically more correct
T(s)=sin,
a s b
 cos — , —
c c' c
s • s bs
a cos — , a sin , —
In the last paragraph on page 59 we take this identification of tangent vectors with
points a little further. We associate the tangent vector Y(s) = (y (s), y 2 (s), y3(s))j3( s )
with the point ( yi (s), y 2 (s), y 3 (s)) of E 3 .
30
M334 II.3
We simply take all the tangent vectors Y(s) and move the point of application j3(s) to
the origin; the new end points of the tangent vectors become the new curve. If we
call this new curve a, then a is parametrized in such a way that a(s ) is the point
with the same Euclidean coordinates as the tangent vector Y(s ). This technique
allows us to transfer certain results about curves to vectors fields on curves, and vice
versa.
Additional Text
The following exercise introduces a result that we shall need in the next section, so
attempt the exercise now. Do not spend too long on it, and be sure to read the
solution.
Page 65, Exercise 7.
Supplementary Comments
(i) Page 57: line 2 The "orthonormal expansion" referred to is that of B' in
terms of T, N, B, as mentioned in Comment (ii).
B' = (B'T)T + (B'N)N + (B'B)B
= (B'N)N if B'T = B'*B = 0.
(ii) Page 59: line 5 Looked at from above we have:
„ „ >.~ cos , a sin— 0)
N(s) = (cosl,sinl, 0)^ A c c
The normal is in the opposite direction to the projection of the position vector 3(s)
on the xyplane, and so it always points towards the zaxis.
M334 II.3 31
(iii) Page 61: line 4 This is not the Taylor approximation to 8 of degree 3: that
would be
0(s) 0(0) + (s Kl s 3 )T + (/c  + ^(0)s 3 )N +k t jB .
We are interested in the approximation of the component of 8(s)  3(0) in each of
the directions T , N , B , so we keep just the term of lowest degree in each
component.
Summary
Notation
T
K
N
B
r
Page 56, line 4
Page 57, line 2
Page 57, line 6
Page 57, line 8
Page 58, line 3
Definitions
(i)
Unit tangent T
(")
Curvature k
(iii)
Principal normal N
(iv)
Binormal B
(v)
Frenet frame field T, N, B
(vi)
Torsion r
(vii)
Frenet approximation
(viii)
Tangent line
(ix)
Osculating plane
(x)
Frenet apparatus T, N, B,
(xi)
Plane curve
(xii)
Spherical curve
Page 56, line 4
Page 57, line 2
Page 57, line 6
Page 57, line 8
Page 57, Lemma 3.1
Page 58, line 3
Page 61, line 5
Page 61, line 10
Page 61, line 14
Page 61, line 12
Page 63, last two lines
of text
Example
(i)
The unitspeed helix
„/ \ i s • s bs
j 3(s) = acos_, asm,—
where c = a + b , has constant curvature a/c and
constant torsion b/c 2 .
Page 58, Example 3.3
(i) The Frenet formulas
T'
=
/cN
N'
=  kT
+ tB
B'
=
tN
(ii) The Frenet approximation: for small s,
Page 58, Theorem 3.2
j8(s)  S(0) + sT(0) + fc(0) N(0) + k(0)t(0) B(0). Page 61, line 4
(iii) A unitspeed curve with positive curvature is a
plane curve if and only if r = 0. Page 61, Corollary 3.5
32 M334 II.3
(iv) A unitspeed curve has zero torsion and constant Page 59, third
curvature k if and only if it is part of a circle with paragraph, and
radius l//c. Page 62, Lemma 3.6
(v) A spherical unitspeed curve has curvature > 1/a, Page 63, last
where a is the radius of the sphere. paragraph
(vi) If a =o:(h) and a are both unitspeed curves, then T
= eT(h), N = N(h), B = e'B(h), k = /c(h), r = r(h),
where e = h' = ± 1. Page 65, Exercise 7
(vii) T = NXB, N = BXT, B = TXN. Text, page 28
(viii) If Y is a vector field on a,
Y = (YT)T + (YN)N + (YB)B. Text, page 29
Techniques
(i) Finding the Frenet apparatus T, N, B, k, t, by dif
ferentiation, followed by taking norms and cross
products. Page 58, Example 3.3
(ii) Identification of a vector field on a curve with Page 59, last
another curve. paragraph
(iii) The following two techniques are useful in
characterizations :
(a) to show a function is constant, show that
its derived function is zero, simplifying the
derived function by using the Frenet
formulas; Page 62, lines 611
(b) prove that "type of curve" => "properties
of k and r" first, as the proof should give
you information which will indicate the Page 62, proof of
method of proving the converse. Lemma II. 3. 6
Exercises
Technique (i)
2. Page 63, Exercise 1. It is sufficient to show that the curve is part of a circle,
and find its centre and radius.
Technique (ii)
3. For j3 as in the preceding exercise, let o be the curve whose coordinate
functions are given by a(s) = T(s). Find the speed of a.
Technique (iii)
4. Page 65, Exercise 10.
M334 II.3 33
Optional Section — Curves in the Plane
5. Curves in the plane are treated somewhat differently from those in E .If
you would like to make a small substudy of them, read the definitions given
in Exercise 8 and then do that exercise. More exercises on curves in the plane
will be provided later. Note that, because E 2 is twodimensional, N spans the
set of vectors orthogonal to T: T' is orthogonal to T because T = constant,
and so T' = /cN for some k.
Solutions
1. Page 65, Exercise 7.
(a) a = <x(h)
so a' = h'a'(h), by Lemma 1.4.5,
so a'(s) = h'(s)i a'(h(s)) for all s.
a is a unitspeed curve, so a'(h(s)) = 1 for all s; S is a unitspeed
curve, so R'(s) = 1 for all s: thus h'(s) = 1 and so h'(s) = ± 1. Since
h is differentiable and has an interval as its domain, h'(s) cannot be
+1 for some values of s and  1 for other values of s. So we can write
h' = e, where e = ± 1 independent of s. Then h(s) = es + h(0) = ± s + s
where s = h(0).
(b) The salient fact from part (a) is that h'(s) = e.
T(s) = a(h)'(s) = h'(s)a'(h(s)), by Lemma 1.4.5,
= cT(h(s)).
f '(s) = eT(h)'(s) = eh'(s)T'{h(s)), by Exercise IL'2.7,
= e 2 T'(h(s)) = T'(h(s)):
thus
7c(s)N(s) = ic(h( S ))N(h(s)): (*)
taking norms we find
5c(s)i N(s) = K(h(s)) HN(h(s))
so jc(s) = k(h(s)) since N(s) = N(h(s)) = 1:
/c(h(s)) and k(s) are positive so we may remove the modulus signs,
obtaining
Jc(s) = K(h(s)).
Now this equation and (*) together imply N(s) = N(h(s)) since
/c(h(s)) is not zero.
B(s) =T(s) XN(s)=eT(h(s)) X N(h(s)) = eB(h(s)).
B'(s) = eB(h)'(s) = eh'(s) B'(h(s)), by Exercise II.2.7,
= c 2 B'(h(s)) = B'(h(s))=r(h(s))N(h(s)):
34
thus
M334 IL3
but
so
?(s).N(s)=T(h(s))N(h(s)),
N(s) = N(h(s))
f(s) = r(h(s)).
The importance of this result is that if a and a are two different
unitspeed reparametrizations (in the sense of Theorem II.2.1) of
some curve 7, then a. and a have the same Frenet apparatus at each
point of the curve.
3.
Page 63, Exercise 1.
4 3
3(s) = (tcos s, 1 sin s, vcos s).
3'(s) = ( t sin s, cos s,^ sin s)
H]3'( s ) II = (f sin 2 s + cos 2 s + ^ sin 2 sf = (cos 2 s + sin 2 s) 2 " = 1 ,
so 3 is indeed a unitspeed curve and
T(s) = 0'(s).
T'(s) = (— r cos s, sin s,jt cos s)
\a q 1
k(s) = T'(s) = (J cos 2 s + sin 2 s +^ cos 2 s)*
25
= (cos s + sin 2 s) 2 = 1
so
N(s) = T'(s).
u 2
U 2
u 3
B(s) = T(s) X N(s) =
4 .
z sins
5
 COSS
3 .
7 sms
5
4
"e COS S
sins
3
•= coss
5
= —r (cos 2 s + sin 2 s),„k (sin s cos s  sin s cos s), — r (sin 2 s + cos 2 s)
1 5' u ' 5/'
B'(s) = so t(s) = 0.
Since r = and /c is constant, Lemma II.3.6 tells us that j3 is part of a
circle of radius— = 1 The proof of Lemma II.3.6 showed that the
centre of the circle is given by ]3(s) + (1/k(s)) N(s), which in this case
gives
4 .3 4 3
(— cos s, 1  sin s, r cos s) + (— r cos s, sin s, r cos s)
= (0,1,0).
a is the curve whose coordinates are the same as those of the vector field T,
and so
lla'(s)T'(s) = K(s).
We saw in the preceding exercise that /c(s) = 1 and so a'(s) = 1 for all s,
that is, o has unit speed.
M334 II.3 35
4. Page 65, Exercise 10.
(a) If a lies on the sphere of centre c and radius r, then
lloe (s)  c = r,
that is,
(a(s) c).(a(s) c) = r 2 .
Using coordinates only and ignoring points of application (following
Comment (iii)), we differentiate to obtain
(a(s)  c) • a'(s) = 0,
that is,
(a(s)  c) • T(s) = 0.
By the orthonormal expansion of a(s)  c in terms of the basis T(s), N(s),
B(s),
a(s)  c = a(s)N(s) + b(s)B(s)
where
a(s) = (a(s)  c)N(s), b(s) = (a(s)  c).B(s).
Differentiating again:
a'(s) = a'(s)N(s) + a(s) N'(s) + b'(s) B(s) + b(s)B'(s),
but
«'(s) = T(s)
so that
T(s) = a'(s)N(s) + a(s)(/c(s)T(s) + r(s)B(s)) + b'(s)B(s)  b(s)r(s)N(s).
Since T(s), N(s), B(s), form a basis, we can equate coefficients on each side:
coefficients of T(s) give
1 =a(s)K(s)
and so
coefficients of N(s) give
= a'(s)  b(s)T(s)
so
Ms) = ^a'(s)
= a(s)p'(s).
Thus a c =  pN p'aB.
Since N and B are orthogonal,
Hc*c 2 =(p) 2 +(p'a) 2
so
r 2 = p 2 + {p'af.
36 M334 II.3
(b) We put 7(s) = a(s) + p(s)N(s) + p'(s)a(s)B(s) as we know by part (a)
that 7(s) is constant when a is a spherical curve. We aim to show that
7(s) is a constant if p 2 + (p'o) 2 = r 2 . We can achieve this if we can
show that 7' = 0.
Now
V = ot' + p'N + pN' + (p'a)'B + p'aB'
= T + p'N + p(/cT + tB) + (p»'B  p'arN
= (1  p/<)T + (p'  p'or)N + (pr + (p'a)')B
= (pr + (p'a)')B since p/c = or = 1.
Now we use our information that p 2 + (p'o) 2 = r 2 . Differentiation gives
2pp' + 2p'a(p'a)' = 0.
so
p + a (p'a)' = 0, because p' # 0,
so
pr + or {p'a)' =
so
pr + (p'a)' = 0, because or = 1.
Thus
7' = (pr + (p'a)')B = and so 7 is constant.
Putting 7(s) = c gives
a(s)  c = p(s)N(s)  p'(s)a(s)B(s)
and so, as before,
lac 2 = (p) 2 + (p'a) 2
= P 2 +(p'o) 2
= r 2 by assumption
so
lac=r,
that is, a lies on the sphere with centre c and radius r.
Page 65, Exercise 8.
(a) N = (y',x'),so
N' = (y",x").
However, T' = (x", y") = kN = »c(y', x') so
x"=k Y '
a 1
y = Kx
and so
N' = (/ex', Ky') =~k{x, y') = kT.
37
Since T has slope angle <£ and unit length
T = cos <p Ux + sin </? U 2
and so
T' = simp <p'Vi + cos«p <p'U 2 .
7T«
N
has slope angle (<p + ■*) and unit length, so
N = cos(<p+)U 1 + sin(<p + )U 2
=  sin^U! + cos<pU 2 .
T' = <p'N.
Thus
However,
T' = kN, and so k = $'.
38 M334 II.4
II.4 ARBITRARYSPEED CURVES
Introduction
This section extends the results of Section II. 3 to arbitrary speed curves. We need to
use the chain rule for reparametrizations of curves (Lemma 1.4.5) and its analogue
for vector fields on reparametrized curves (Exercise II.2.7) to adapt the Frenet
apparatus to an arbitrary speed curve. We are searching for an effective procedure
for computing the Frenet apparatus of an arbitrary regular curve, and, although we
succeed in this, we lose the simplicity of the earlier Frenet formulas, with the result
that the theory becomes cumbersome and the computations more involved and so
liable to error. Although this whole section is important and very practical, if you
find yourself becoming bogged down in calculus and computation give it a rest for a
while and return later: the results of this section are not needed again until Section
III.5.
READ: Section II.4 (pages 6674), omitting the last full paragraph on page 67.
Comments
(i) Page 66: the definitions The curvature, torsion etc. of a are those at the same
point of the corresponding unitspeed curve a", that is, if p is the point <x(t) =o?(s(t))
on the curve, then the curvature at p is K(t) = lc(s(t)). There is a possible difficulty
here, in that if we take different unitspeed reparametrizations of a. they may give a
different Frenet apparatus. However, the result of Exercise II. 3. 7 tells us that so
long as we restrict ourselves to unitspeed reparametrizations in the sense of
Theorem II. 2.1 the Frenet apparatus is the same. Thus "the Frenet apparatus of a is
the Frenet apparatus of any unitspeed reparametrization of a" is a valid definition.
(it) Page 73: the orthonormal expansion ofu By Theorem II. 1.5
u = (uT)T + (uN)N + (uB)B
= cos # T + aB for some a, because u*T = cos # and u«N = 0.
Now u is a unit vector, so
l = u 2 = cos 2 # + a 2
so
so
a 2 = 1  cos 2 # = sin 2 #
a = ± sin #.
If a = + sin # we have the required expression; if a =  sin # then, because cos ( #) =
cos # and sin (#)= sin #, we can replace # by & and obtain
u = cos(#)T + sin(#)B.
This is possible because we are allowing # to have any value. The recognition that we
can just write sin # for a in such a situation should become second nature to you.
M334 II.4
Supplementary Comments
39
(i) Page 67: second paragraph You are not seriously expected to
reparametrize the curve a(t) = (t, t 2 , t 3 ). If you attempted it and got stuck
then you will understand O'Neill's point — there may not be explicit
formulas for a.
(ii) Page 69: the formula for K Since v = a', the equation
(iii)
yields
la X a" = kv 3
la X a
la X dl
a
Ml 3
Page 71: the example
i
i z
^^^x ^
I y "
/ \ b
/ \ c
"J
**x
Summary
Notation
a
Definitions
(i) Frenet apparatus for an arbitraryspeed curve
(ii) Spherical image o
(iii) Cylindrical helix
(iv) Circular helix
Results
(i) T' = /cvN
N'=kvT + tvB
B' =  rvN
Page 71, line 10
Page 66, first paragraph
Page 71, line 10
Page 72, Definition 4.5
Text: page 41, Exercise 5
Page 67, Lemma 4.1
40
(")
(iii)
(iv)
a =tT + kv 2 N
M334 II.4
Page 68, Lemma 4.2
T =
B =
a' X a'
a' X a
\a X a
N = B X T
a
r =
'3
(?L2L£Lk°L
lla' X a"ll 2
For a regular curve:
K = if and only
t = if and only
K constant >0, r = if and only
K constant >0, r constant =£0 if and only
t/k constant if and only
Techniques
(i) Finding the Frenet apparatus of an arbitrary
regular curve.
(ii) Finding the spherical image of a curve.
(iii) Finding the Frenet apparatus for a curve derived
from some other curve in terms of the latter's
Frenet apparatus: this involves repeated
application of the Frenet formulas and taking
higher derivatives.
(iv) Finding u and # for a cylindrical helix by putting
cot & = t/k and u = cos # T + sin # B.
Page 69, Theorem 4.3
if a is a straight line
if a is planar
if a. is part of a circle
if o: is a circular helix
if a is a cylindrical helix.
Page 74
Page 69, Theorem 4.3
Page 71, second paragraph
Page 71, last paragraph
Page 73, Theorem 4.6
Exercises
Technique (i)
1. Page 74, Exercise 2. This continues the example of Exercise II.2.3, whose
solution is on Text page 23.
Techniques (ii) and (iii)
2. Page 75, Exercise 11. K a has been found on page 72, so all you need find is
t . Make use of the working already done on pages 7172.
Technique (iii)
3. Page 74, Exercise 8.
Technique (iv)
4. Page 75, Exercise 9(c).
M334 II.4 41
Theory Exercises (omit if you are short of time)
5. Page 75, Exercise 10. This continues from Exercise II.4.8: a cylindrical helix
is a circular helix if its crosssection curve y is part of a circle.
6. Page 75, Exercise 12(a). This continues from Exercise II.4.11.
Optional Section — Curves in the Plane
If you have time to follow this optional line of study, you should also do:
7. Let a be a plane curve, a a unitspeed reparametrization of a, that is a = a(s)
where s is an arclength function for a. Define
T = T(s)
K = k(s)
N as in Exercise 8 on page 65.
Prove that (a) a' = vT
(b) T' = vkN
(c)N' = kvT.
8. Page 75, Exercise 14.
Page 75, Exercise 15. The central curve is defined in Exercise II.4.13. After
you have sketched both curves, how do you account for the cusp (sharp
point) in a* at a* (0)?
Solutions
1. Page 74, Exercise 2.
a(t) = (cosh t, sinh t, t).
We already know that:
a'(t) = (sinht, cosh t, 1)
v(t) = ^2 cosht
s(t) = N /2sinht.
Now
a"(t) = (cosh t, sinh t, 0)
a'"(t) = (sinht, cosht, 0)
so a'(t) X a"(t) = (sinh t, cosh t, sinh 2 1 cosh 2 1) = (sinh t, cosh t,  1),
oc'(t) X a"(t) = (sinh 2 t + cosh 2 t + 1)2 =yj2 cosht,
c/(t) X a"(t).a'"(t) = sinh 2 t + cosh 2 t = 1.
42 M334 II.4
Now we simply apply Theorem II.4.3:
T(t) = W^\ = 72 (tanh tj h sech t} (writing tanh for ^ and sech for ^
B W " a'(tj X a"(t) J2 ( " tanht ' ^~ secht )
N(t) = B(t) X T(t) = (2secht, 0, 2tanht) = (secht, 0, tanht)
,., n _ ll«'(t)Xa"(t)ll _ V2cosht _ 1
{ } Ha'(t) 3 (V2 cosht)* 2cosh 2 t
T W " a'(t) X a"(t)] 2 " (V2 cosh t) 2 2 cosh 2 t
Since we know s(t) =y/2 sinht, 2cosh 2 t = 2 + 2sinh 2 t = 2 + s 2 ,
soic(s) = T(s) = ^p.
Page 75, Exercise 11.
Let j3 have Frenet apparatus T, N, B, K, r.
Making the usual identification of coordinates and ignoring both the
distinction between points and tangent vectors and the points of application
of tangent vectors, we can write, as on pages 7172:
a = T
a = fcN
a" = k 2 T + k'N + ktB
a'x o" = k 2 (kB + tT)
\<j' xo"\\ = K 2 (K 2 + T 2 )\
Now
so
a'" =  2kk'T  k 2 T' + k"N + k'N' + k'tB + kt'B + ktB'
=  2kk'T  k 3 N + k"N  k'kT + k'tB + k'tB + kt'B  kt 2 N
=  3/c/c'T + (k"~ k 3  kt 2 )N + (kt + 2k't)B
a' X a" a'" = k 2 t( 3kk') + k 3 (kt' + 2k't)
= k 3 (kt'  tk)
= k 5 (tIk)\
_ o'Xo"'o" , _ k 5 (t!k)' _ (t/k)'
T ° lia'Xa"!! 2 k*(k 2 +t 2 ) k(1 + (t/k) 2 )
3. Page 74, Exercise 8.
(a) We have to show that f(t)' = (7(t)  a(0))u = 0.
We know that
f(0) = (7(0)  a(0))u = s(0) cos # uu =
M334 II.4 43
because s(0) = 0, and so it is sufficient to show that f is a constant
function, i.e. that f = 0.
f(t) = ( 7 (t)a(0)).u
= (a(t)  s(t)cos du a(0))u
= (a(t)  a(0))u  s(t)cos # because uu = 1 ;
f'(t) =a'(t).u s'(t)costf
= v(t)T(t)u v(t)cos#
= v(t)cos &  v(t)cos #
= 0.
Hence 7(t)  a(0) is always perpendicular to u, that is, y lies in the plane
through a(Q) orthogonal to u.
(b) Let us suppose that a has unitspeed, so that s(t) = t and v(t) = 1.
Using bars to denote the Frenet apparatus of 7:
7(t) = a(t)  tcos # u
7'(t) = a'(t)  cos # u = T(t)  cos # u. (*)
v=H7'll
* ((T  cos & u).(T  cos # u))2
= (TT  2 cos & Tu + cos 2 & uu)2
= (1 2 cos 2 »? + cos 2 #p
= (1 COS 2 #)2
1
= (sin 2 ^)2" = sin^.
Since v is constant, Lemma II.4.2 gives:
7 "(t) = K(t)(v(t)) 2 N(t)
= K(t) sin 2 # N(t),
44 M334 II.4
but our first expression, (*), for 7'(t) gives:
T "(t) = T'(t) = K(t)N(t),
so£(t)sin 2 #N(t) = /c(t)N(t).
Taking norms gives K(t) = (K(t)/sin 2 #) since ic(t),7c(t), and sin 2 # are
all positive.
4. Page 75, Exercise 9(c).
We use the technique used in the proof of Theorem II.4.6. We already have
all the necessary information about a (see solution to Exercise II.4.2).
r(t) = /c(t) = j, so r(t)/fc(t) = 1: we choose
7r r(t)
# = so that cot# = 1 = — x .
4 K(t)
Now we put U(t) = cos  T(t) + sin  B(t)
= _1.J_ (tanht, l,secht)+— — (tanht, 1,secht)
V2 V2 V2 V2
= (0,1,0).
This vector field U is parallel, as expected, so we put u = (0, 1, 0), and u is a
fixed unit vector such that T(t)u = cos (7r/4).
7T
7 (t) =a(t) s(t)cosu
= (cosht, sinht.t)^ sinht. — (0, 1, 0)
= (cosht, 0, t).
7(t)  a(0) = (cosh t, 0, t)  (1, 0, 0) = (cosh t  1, 0, t)
so(7(t) a(0)).u = (cosht 1, 0, t)(0, 1, 0) = 0, as required.
5. Page 75, Exercise 10.
Since K > and T # are constants, we can apply Theorem II.4.6: thus /3 is
certainly a cylindrical helix with
# = cof 1 (j\k\ u = cost? T(t) + sintf B(t).
All that remains to show is that the crosssection curve 7 of is part of a
circle. By Exercise 11.4.8(a), 7 is planar; by Exercise 11.4.8(b) the curvature
of 7 is (K/sin 2 #), which is a constant: hence, by Lemma II.3.6 (which is
unaltered if the unitspeed restriction is lifted), 7 is part of a circle.
Page 75, Exercise 12(a).
We may suppose the curve 8 has unitspeed. Let a be its spherical image.
8 is a cylindrical helix <=> t/k is constant (by Theorem II.4.6)
<=> k is constant and r ff is zero (by Exercise II.4.11)
<=» a is part of a circle (by Lemma II.3.6 and its converse).
M334 II.4 45
7. (a) a = 5(s)
so
a. = s a (s), by Lemma L4. 5,
= vd'(s), because s' = v,
= vT(s)
= vT.
(b) T = T(s)
so
if tmt
T' = s'T'(s), by Exercise II.2.7,
= vT'(s)
= v/c(s)N(s), because T' = kN,
= vkN.
(c) N = N(s)
so
r' 'tvt'
N' = s'N'(s) , by Exercise II.2.7,
= vfl'(s)
= v(Jc(s)T(s)), by Exercise II.3.8,
= v/cT.
8. Page 75, Exercise 14.
c/ = vT
a" = v'T + vT'
= v'T + v 2 kN
Thusa".N = v 2 /c.
However, in the plane N = J(T) = — , so
v
 ,,2
_ a".J(c*')
/c =
v 3
In terms of the coordinate functions,
«' = ( x '> y')«
](«') =(y',x') a i
v = (x' 2 +y' 2 j5
a  ( x » y )«
x y  x y
so /c = ,
(x' 2 +y' 2 )t
46
9.
M334 II.4
Page 75, Exercise 15.
(a)
VK K
= vT   T   N
K K l
K
=  2 N.
The tangent line to a* at a*(t) is the line through a*(t) in the direction
a*'(t); the normal line to a at a(t) is the line through a(t) in the direction
N(t), that is in the direction orthogonal to a' (t). Consequently these two
lines are the same if and only if
(i) a*'(t) and o/(t) are orthongonal: a*'(t)a'(t) =
(ii) there is a number u(t) such that a*(t) = a(t) + u(t)N(t).
«'(t)
a*'(t)
a*(t)
Firstly, let us check that a* does satisfy these conditions.
a *'(t).a^t)="^ 2 N(t).a'(t) =
so (i) is satisfied. Moreover
a*(t) = a(t) + — N(t)
K(t)
I
so condition (ii) is satisfied with u(t) = — .
Conversely, suppose )3 is a curve satisfying (i) and (ii), that
is
and
ff.a' =
= a + uN.
M334 II.4 47
Then
P' = ol' + u'N + uN'
= vT + u'N  uv/cT
= v(l  uk)T + u'N
so
ffd = (v(l  u/c)T + u'N).(vT)
= v 2 (l UK).
j3'«o:' = and v is nonzero, so this implies K = — , that is
u
1
j3 = a + — N = a*.
T(a') a"T(a')
(b) We have already shown that N = ±J , K = — ^— ',
v v~
1
so a* = a + — N
K
v J J(a')
= a + JV '
a JK) v
a' •a'
a .J(a)
(c) a(t) = (t + sin t, 1 + cos t) TT<t<ir
a'(t) = (1 +cost, sint)
a"(t) = (sint, cost)
c/(t)a'(t) = 1 + 2 cost + cos 2 t + sin 2 t = 2 + 2 cost
J(a'(t)) = (sin t, 1 + cos t)
a"(t)J(a'(t)) =  sin 2 1  cos t  cos 2 1 =  1  cos t.
*/ v / v a '(t)a'(t)
Now a*(t) = a(t) + n^ — — ](a'(t))
U «"(t)'J(«'(t)) Jl W '
= a(t)2J(cx'(t))
= (t + sin t, 1 + cos t)  2(sin t, 1 + cos t)
= (t  sin t,  1  cos t).
48
M334 II.4
The cusp at a*(0) suggests that a* is not regular at that point. In fact
that is the case, for
o:*'(t) = (1  cost, sint)
so
a*'(0) = (0, 0).
M334 II.5 49
II.5 COVARIANT DERIVATIVES
Introduction
This section depends on Sections 1.3 and II. 2; you also need to be familiar with the
differential of a function from E 3 to R (Section 1.5). We introduce a new derivative,
the covariant derivative. This is a very important idea in differential geometry, and
serves as the starting point for a lot of more advanced work. The definition is similar
to the other definitions of derivatives in the book, in that it is given in terms of a
derivative we already know along the particular curve ti >p + tv. It generalizes the
derivative of a vector field on a curve.
READ: Section II. 5 (pages 7780) .
Comment
(i) Page 78: the symbol V This is usually pronounced "del" or "nabla".
Additional Text
(i) Covariant Differential If W = SwjUi is a vector field on E 3 , the covariant
differential, VW, of W is the function on tangent vectors defined by
VW(v) = V V W.
By Lemma II.5.2 V V W = Zv[wj] U^p) and so
VWM^ZvhrJUifo)
= 2dwi(v)Ui(p), by definition of dwj,
= (ZdwiUi)(v).
Thus we may write
VW = SdwiUi
(n) A Useful Result The result of the following exercise will be used often
later in the course. Attempt the exercise now.
1. Page 81, Exercise 6(a). (HINT: Use Lemma 1.4.6.)
Summary
Notation
V v w Page 78, line 2
V V W Page 79, line 15
V ™ Text, page 49
50
Definitions
(i) Covariant derivative with respect to a tangent
vector V V W
M334 II.5
Page 77, Definition 5.1
(ii) Covariant derivative with respect to a vector field
VyW Page 79, line 12
(iii) Covariant differential VW
Text, page 49
Results
(i)
(i")
(iv)
If W = 2 wjUj, V V W = 2 v[wj] Ui(p)
V v W = 2V[w i ]U i .
Linearity and Leibnizian properties of the
covariant derivative.
With respect to a tangent vector v:
V a v + bwY = aV v Y + bV w Y
V v (aY + bZ) = aV v Y + bV v Z
V v (fY)=v[f]Y( P ) + f(p)V v Y
v[Y'Z] = V v Y'Z(p) + Y(p)'V v Z
With respect to a vector field V:
V fV + gW Y = fV vY + gVwY
V V (aY + bZ) = aV V Y + bV V Z
V v (fY) = V[f]Y + rV V Y
V[YZ] =V V YZ + YV V Z
If W = 2 wjUj, VW = 2 dwiUj.
V(t)W = (w(«))'(t).
Page 78, Lemma 5.2
Page 79, line 18
Page 78, Theorem 5.3
Page 80, Corollary 5.4
Text, page 49
Page 81, Exercise 6(a)
Techniques
(i) Finding the covariant derivative with respect to a
tangent vector or vector field,
(a)
(b)
(c)
(d)
from first principles; Page 77, Definition 5.1
using result (i) and calculating the
directional derivatives from first principles; Page 78, Lemma 5.2
using result (i), the linearity and Leibnizian
properties of both the directional and
covariant derivative, and the results
Ui[f] =(3f/3xi),v[f] =Svi(8f/axi);
by means of the covariant differential. Text, page 49
Page 78, Lemma 5.2,
Page 78, Theorem 5.3 and
Page 80, Corollary 5.4
M334 II.5
Exercises
Technique (i)(a)
2. Page 80, Exercise 1(a). Work directly from Definition II.5.1.
51
Technique (i)(b)
Page 80, Exercise 1(a). Use the result of Lemma II.5.2 but calculate the
directional derivatives from first principles.
Technique (i)(c)
4. Page 80, Exercise 1(a). Now use all the results available to simplify the work!
5. Page 80, Exercise 2(a).
6. Page 80, Exercise 2(c).
Technique (i)(d)
7. Page 80, Exercise 5.
Solutions
Page 81, Exercise 6(a).
LetW = 2w i U i .
Then V (t) W = ?a'(t) [wj] Ui(a(t)),
1 d(w:(a))
i dt
= (?^(a(t))Ui( a (t)))',
i
= (W a )'(t).
by Lemma II. 5. 2,
by Lemma 1.4.6,
by the definition of differentia
tion of a vector field on a curve,
Page 80, Exercise 1(a).
p = (l,3,l),v = (l,l,2),so
p + tv = (l+t, 3 t,l + 2t).
W = x 2 U!+yU 2
so W(p + tv) = (1 + t) 2 Uj (p + tv) + (3 t)U 2 (p + tv).
V v W = W(p + tv)'(0)
= [2(l+t)U 1 (p + tv)U 2 ( P + tv)] t =
= 2U,(p)U 2 (p)
= (2,l,0) p .
52 M334 II.5
3. Page 80, Exercise 1(a).
W = x 2 U! + yU 2
so
V V W = v[x 2 ] U^p) + v[y] U 2 (p) by Lemma II.5.2.
v[x 2 ] =x 2 (p + tv) t =
at
J t d + t)1 t .o
= 2(l+t) t =
= 2;
v[y]'= Ty(p + tv )t = o
at
f t (3t) t ,
= it = o
= 1.
Thus
V V W = 2U 1 ( P )  U 2 (p)
= (2,l,0) p .
Page 80, Exercise 1(a).
V v W = v[x 2 ]U 1 (p)+v[y]U 2 (p).
3(x 2 ) 3(x 2 ) 3(x 2 )
V X 2 =V!r — (P)+V 2 — (p) + V 3 — — (p)
ox dy dz
= Vl 2x(p)
=1.2.1 = 2;
3y dy dy
v [y] = v i r~(p) + v 2 r(p) + v 3 — (p)
ox oy dz
= v 2
= 1,
so
V v W = 2U 1 (p)U 2 (p)
= (2,l,0) p .
M334 II.5 53
5. Page 80, Exercise 2(a).
W = cos xUj + sin x U 2
so
V V W = V[cos x] U 2 + V[sin x] U 2 by Lemma II.5.2.
V[cos x] = (yUi + xU 3 ) [cos x]
= yU 1 [cosx] +xU 3 [cosx], by linearity,
9(cos x) 9(cos x)
= y + x
OX 8z
= y sinx + 0;
V[sin x] = (yUi + xU 3 )[sinx]
= yUJsin x] + xU 3 [sin x] , by linearity,
9(sin x) 3(sin x)
= y 1 x
3x 3z
= y cos x + 0.
Thus
VyW = y sin x U l  y cos x U 2 .
6. Page 80, Exercise 2(c).
V v (z 2 W) = V[z 2 ] W + z 2 V V W by Corollary II.5.4.
V[z 2 ] =2zV[z], by Corollary 1.3.4,
= 2z(yU 1 + xU 3 )[z]
= 2yz U x [z] + 2xz U 3 [z] , by linearity,
= 2yz.0 + 2xz.l
= 2xz,
so
V v (z 2 W) = 2xz W + z 2 V v W
= 2xz (cos x Ui + sin x U 2 ) + z 2 (y sin xUjy cos x U 2 )
= (2xz cosx + yz^inxjUj + (2xz sin x  yz 2 cos x)U 2 .
7. Page 80, Exercise 5.
W = xy 3 U!  x 2 z 2 U 3
so
VW = d(xy 3 )U 1  d(x 2 z 2 )U 3
= (y 3 dx + 3xy 2 dy)U!  (2xz 2 dx + 2x 2 z dz)U 3 .
(a) (y 3 dx + 3xy 2 dy) ((1, 0, 3)(_ 1} 2 , 1)) = 2 3 .1 + 3.(l).2 2 .0 = 8
(2xz 2 dx + 2x 2 zdz) ((1, 0, 3) ( _ ^ 2> _ 1} ) = 2.( 1).( 1) 2 .1 + 2.( 1) 2 .( l).(3) = 4
and so
VW ((1, 0, 3) ( _ 1? 2 , l)) = SU^p)  4U 3 (p)
= (8,0,4) p .
54 M334 II.5
(b) (y 3 dx + 3xy 2 dy) ((1, 2, l) (lf 3j 2 )) = 3 3 .(l) + 3.1.3 2 .2 = 27
(2xz 2 dx + 2x 2 z dz) ((1, 2, 1) (1> 3, 2 )) = 2.1.2 2 (1) + 2.1 2 .2.(1) = 12
and so
VW((1, 2, l)(i, 3, 2)) = 27U 1 ( P ) + 12U 3 (p)
= (27, 0, 12) p .
M334 II.6 55
II.6 FRAME FIELDS
Introduction
This section does not depend on the previous two, but on Sections II. 1 and II.3. The
idea of a general frame field (as opposed to the Frenet frame field of Section II.3) is
formulated.
READ: Section II.6 (pages 8184) .
Additional Text
(i) The Pointwise Principle In this section O'Neill has used the pointwise
principle to extend some familiar definitions and theorems from the set of
all tangent vectors to E 3 to the set S of all vector fields V in E 3 . We recall
that S is only a threedimensional module (see Comment (ii) to Section 1.2
(Text, page 10)) and so S is not isomorphic to E 3 . However, because each
element of S can be expressed as 2f^U, many of the ideas do carry over to S
via the pointwise principle with no trouble. Some examples are given below.
(a) The vector fields E u E 2 , E 3 form a basis for S if the tangent vectors
E i (P)> E 2(p), Ea(p) form a basis for T (E 3 ) for each point p of E 3 .
(b) The dot and cross products and norm are defined pointwise. That is,
if V and W are vector fields then V«W is the function from E 3 to R
defined by
(V.W)(p) = V(p).W(p),
V X W is the vector field defined by
(VX W)(p)=V(p)X W(p),
and V is the function from E 3 to R defined by
IIVII(p) = HV(p).
(c) Vector fields V and W are said to be orthogonal if V(p)W(p) = for
all p. For example, V X W is orthogonal to V and W.
(d) Similarly the vector fields E t , E 2 , E 3 are orthonormal if
that is (E i E i )(p) = 1 for all p and for i =■ 1, 2, 3 and Ej, Ej are
orthogonal for i ¥= j.
(e) (Lemma 6.3). If Ej , E 2 , E 3 are orthonormal vector fields, any vector
field V has an orthonormal expansion of the form
V= (V.E,)^ + (VE 2 )E 2 + (V.E 3 )E 3
= f 1 E 1 + f 2 E 2 + f 3 E 3
where fj = V«Ej.
Moreover, if W = gjEj + g 2 E 2 + g 3 E 3 , we find, as one would expect, that
VW=f lgl +f 2g2 +f 3g3 .
In future we shall not always trouble to point out when the pointwise
principle is used to make definitions or deduce results about vector fields. So
long as you are sufficiently familiar with the linear algebra of E 3 you should
have no trouble with adapting this to vector fields.
56
Summary
Notation
VW
vx w
livii
E t , E 2 , E 3
r, #, z
p, #, <£
M334 II.6
Page 82, line 1
Text, page 55
Page 82, line 2
Page 82, Definition 6.1
Page 82, Example 6.2.(1)
Page 83, Example 6.2.(2)
Definitions
(i) Orthogonal vector fields
(ii) Orthonormal vector fields
(iii) Frame field E l5 E 2 , E 3
(iv) Coordinate functions with respect to a frame field
Text, page 55
Text, page 55
Page 82, Definition 6.1
Page 83, Lemma 6.3
Examples
(i) The natural frame field U l5 U 2 , U 3
(ii) The cylindrical frame field Ej, E 2 , E 3
(iii) The spherical frame field F lf F 2 , F 3
Page 82, second paragraph
Page 82, Example 6.2.(1)
Page 83, Example 6.2.(2)
Page 82, Example 6.2.(1)
Results
(i) For the cylindrical frame field
E x = cos# l^ + sintf U 2
E 2 =  sin# l^ + costf U 2
E 3 = U 3 .
(ii) For the spherical frame field
F t = cosip cos# L 1 ! + cos<p sin# U 2 + sirup U 3
F 2 = sin# Ui i cost? U 2
F 3 = sin</?cos# L 1 !  sin<£sin# U 2 + cos</>U 3 .
Page 83, Example 6.2.(2)
(iii) If E l5 E 2 , E 3 is a frame field and V is a vector field,
then
V = (VE^E! + (VE 2 )E 2 + (VE 3 )E 3 . Page 83, Lemma 6.3.(1)
(iv) If E l5 E 2 , E 3 is a frame field, V = SfjEi and
W = SgjE, then
V.W = Sfg. Page 83, Lemma 6.3.(2)
Techniques
(i)
Page 82, Definition 6.1
Recognition of frame fields,
(ii) Production of frame fields using GramSchmidt and
cross product (see Lemma ILL A).
(iii) Expression of a vector field in terms of a frame
field, and applications thereof. Page 82, Example 6.2
M334 II.6 57
Exercises
Techniques (i) and (ii)
1. Page 84, Exercise 1.
Technique (Hi)
2. Page 84, Exercise 2.
Solutions
w (p)* Lj.r v (p>
Page 84, Exercise 1.
First we notice that V and jW are never zero and so E 1 and E 2 are
welldefined. Since V and W are linearly independent at each p, we know
that
V(p) *
W(p) ¥=
(WV)(p )
IIV(p) 5
V(p) =£ and so V is never zero.
W = W (WE^Ej
IIVH 2
which is also never zero.
We must show that EE: = 5.
v_jv__ y^ _ iivii 2
livifiivii" livil 2 "hvii 3
E.E, = — • — = = — — = i •
W W
E2 * E2 = Mi' Mi ^ similarly '
E . E  = _X_ . _W_ 3 VW = V(W WEtEt)
r 2 " IIVH W = iivii W IIVH iiwii
_ VW (WE^fVE,) _ V(E r W)  (WEOfllVHErEQ
IIVH W hvii iiwii
IIVH (WEQ  [V(WE 1 ) _
IIVH iiwii
Since E 3 = E 1 X E 2 , £^£3 = E 2 «E 3 = and
E 3 E 3 =((E 1 E 1 )(E 2 .E 2 )(E 1 .E 2 ) 2 ) 2 , by Lemma II. 1.8,
= 1.
Page 84, Exercise 2.
(a) Ux = cos# E t  sin# E 2
= cosip cos?? Fj  sin# F 2  sirup cost? F 3 .
(b) cost? Uj + sin# U 2 + U 3 = Ej + E 3
= (cos<p + sin^F^ + (cos<£  sin</?)F 3
58 M334 II.6
(c) (i) xUi + yU 2 + zU 3 = r cost? Ux + r sint? U 2 + zU 3
= rE, + z U 3 ;
(ii) xUx + yU 2 + zU 3 = p cos<p cost? Ui+ p cosy sint? U 2 + p simp U 3
= pFi.
If you cannot do this type of exercise by inspection, you may find the
following method helpful. We can write the equations giving the Fj in terms
of the Uj in matrix form as
Fj \ /cos <p cost? cos <p sin t? simp \ lUi
F 2 I =
F 3
 sint? cost? .
 sirup cost?  simp sint? cos<p
U 2
= M
u 2
u 3
Since the Fj and the Uj are both frames, M must be an orthogonal matrix,
i.e. M
1 = t M, so we can write /U/
U 2
\U 3 ,
= t M IF A , giving the Uj in terms of the F.
F 2
As an example, we work part (b)(ii):
cost? Ui + sint? U 2 + U 3 = (cost?, sint?, 1)
Ui\
u 2
u 3
= (cost?, sint?, l)tM /Fi
F 2
h
= (cost?, sint?, 1) / cos<p cost? sin t?  simp cost?
cos<p sin t? cos i?  sirup sin t?
sirup cos<p
= (cosip + simp, 0, cos<p  sin<p) I F x \
F 2
\
Fi
F 2
F 3
= (cosip + simp)Fx + (cos<p  sin<p)F 3
M334 II.7 59
H.7 CONNECTION FORMS
Introduction
These last two sections of Chapter II are of quite a different nature to the preceding
ones. The material is more abstract, and concerned almost entirely with forms. We
expect you will find these sections difficult. However, they are not needed again
until Chapter VI, so you may be content with just a superficial understanding at this
stage of the course. If necessary, you can revise these sections when you reach
Chapter VI: you should find them easier going at that reading after the lapse of
time.
Section II. 7 draws together the previous two sections and Section 1.5. Its purpose is
to generalize the Frenet formulas to arbitrary frame fields. The matrix techniques
introduced look messy but are simple to use in practice.
READ: Section II. 7 (pages 8590).
Comment
(i) Page 88, Theorem 7.3 The ijth entry of dA*A = 2(dA)., (*A) . =
k *k kj
2 (dajj c )(a:j c ), which we write as Sa^da^ to keep to the standard con
IV IV
vention of putting the differentials on the right. We cannot use such a
conventional arrangement with the matrices dA and t A, however, as matrix
multiplication is not commutative.
Supplementary Comment
(i) Page 89: dA The fact that d(cos #) = sin § d#, and the other similar
results, follow from Lemma 1.5.7.
Summary
Notation
"ij Page 85, Lemma 7.1
w Page 87, line 4
A Page 88, line 11
dA Page 88, line 18
Definitions
(i) Connection forms c^j Page 85, Lemma 7.1
(ii) Attitude matrix (A) of a frame field Page 88, line 11
(iii) Differential of a matrix whose entries are functions
from E 3 to R Page 88, line 18
Results
(i) cojj is a 1form and cjjj =wji. Page 85, Lemma 7.1
(ii) The connection equations: V^ = Scojj (V)Ej. Page 86, Theorem 7.2
(iii) 03 = dA*A. p age 88, Theorem 7.3
Technique
(i) Calculation of the connection forms using 03 = dA^.
Page 88, Theorem 7.3
60
Exercises
M334 II.7
Technique (i)
1. Page 91, Exercise 4, using A and co.
Theory Exercise
2. Page 91, Exercise 5.
Solutions
1. Page 91, Exercise 4.
A= j cos <p cost? cos <p sin # sin<p
sin i? cos#
,sintp cost? simp sin# cos<p
dA =
cos <p d<p \
sin<pd<p /
 sin <p cos # d<p  cos <p sin # d#  sin<p sin i? dtp + cos <p cos # d#
cos#d# sin#d#
'  cos <p cos # dip + sin <p sin # d#  cos <p sin # dy?  sin <p cos # d#
co = dAtA =  cos<p d& d<p \
cos<pd# sin<pd#
dip simp d#
so oj 12 = cos<p d#
co 13 = d<p
co 2 3 = sin<p d#.
2. Page 91, Exercise 5.
i
= 2 V v (f i E i ), by Corollary 11.5.4.(1),
i
= 2 (Vtfj] Ej + fjVyEi), by Corollary 11.5.4.(3),
i
= 2 (V[fj] Ej + fj? co(V)E), by the connection equations,
i J
= 2 V[f:]E:+S 2 f i w i :(V)E:, by rearrangement,
i J J j i J J
= 2 {V[fj] + 2 f i co ij (V)}E j .
In the case of the natural frame field all the connection forms are zero (prove!)
and so this equation reduces to
V V W = 2V[fj]Uj
as given on page 79.
M334 II.8 61
II.8 THE STRUCTURAL EQUATIONS
Introduction
As mentioned in the previous section, you will probably find this section rather
abstract and hard going. Do not attempt it until you feel fairly confident about both
Section II.6 and Section II. 7. You should also be familiar with the ideas of linear
functional, dual space and dual basis (see Unit M201 12, Linear Functional and
Duality, Sections 1 and 2).
This section deals with finding the exterior derivatives of the connection forms
introduced in Section II. 7 and of the dual 1forms of a frame field. As we do not yet
know what an exterior derivative is, this section, like 1.6, must be approached purely
formally, as manipulation of strings of symbols. In Chapter IV we shall find a
meaning for the exterior derivative, and in Chapter VI we shall relate that to the
results developed here.
The results of this section are not needed until Chapter VI. You will probably find it
hard: if so, do not worry about understanding it in depth now, but return to it from
time to time so that it becomes fairly familiar before Chapter VI is reached.
READ: Section II.8 (pages 9195).
Comments
(i) Page 91: dual 1forms In Lemma II.6.3 we saw that if Ej , E 2 , E 3 is a
frame field then any vector field V can be expressed as
V = (VEJE! + (V.E 2 )E 2 + (VE 3 )E 3 .
In a similar manner, we obtain the fact that, if v is a tangent vector,
V P = ( v p Ei(p))Ei(p) + (v p E 2 ( P ))E 2 (p) + (v p .E 3 (p))E 3 (p),
simply by applying Lemma II.1.5 to the frame E x (p), E 2 (p), E 3 (p). Thus if
we define the functions 0j from the set of all tangent vectors to E 3 to R (and
extend to vector fields by the pointwise principle) by
i(v p ) = Vp.Ei(p)
we may write
v p = MvpJE^p) + 2 (v p )E 2 ( P ) + 3 (v p )E 3 (p)
V = 0,(V)E, + 2 (V)E 2 + 3 (V)E 3 .
0i is certainly linear at each point, and so 0j is a 1form. In fact, since the Ej
are orthonormal, we have
«i(Ej) = «ij
which is precisely what is meant by saying that the 0j are the dual basis to
the Ej. This is why we call the 0; the dual 1forms of the frame field.
(ii) Page 94: Example The 0; have not been found yet, but a method for so
doing is given on page 95. The cojj were computed in Exercise II.7.4.
62
Summary
Notation
d:
M334 II.8
Page 91, Definition 8.1
Definition
(i) Dual 1form { .
Page 91, Definition 8.1
Results
(i)
(ii)
(iii)
(iv)
(v)
If V is a vector field, then V = S0 i (V)E i .
*i(Ej) = 5 ij.
If is a 1form, then = S0(E i )0j.
If Ej = S ajU., then 0j = 2 ajj dxj.
J J
The structural equations
Techniques
(i) Finding the dual 1 forms 0} by adhoc methods
or using the attitude matrix.
(ii) Using the structural equations.
Text, page 6 1
Text, page 6 1
Page 92, Lemma 8.2
Page 92, line 20
Page 92, Theorem 8.3
Page 92, line 20
Page 92, Theorem 8.3
Exercises
Technique (i)
1. Page 96, Exercise 4. Part (c) depends on knowing what 9f/dr, 9f/8# mean
and on knowing the formula
3f 3f n 3f
df = dr + — d# +— dz.
9r o# oz
These will be explained in the final section of this text, so you may like to
defer this part of the exercise until then.
Technique (ii)
2.
Check the structural equations for the cylindrical frame field. (The con
nection forms were calculated on pages 8990.)
Theory Exercise
3. Page 95, Exercise 1.
M334 II.8
Solutions
1. Page 96, Exercise 4.
(a) E x = cost? L^ +sin# U 2
E. 2 =sint? JJi + cost? U 2
E 3 = U 3 .
63
By definition, ^(U:) = UEi = E^IL, so
1 (U 1 ) = cost> 1 (U 2 )=sint? 0i(U 3 ) = O.
We compare this with the value of dr on each of the U.
dr_
>X:
dr(U i ) = U i [r] =
1 l o:
by Lemma 1.3.2.
2\2
SO
Thus
r = (x 2 + y 2 )
dx ( x 2 +y 2 )2 r
= — = cost?
3r
9y (x 2 +y 2 ) 2 r
L==sin#
dr
— = 0.
3z
dr(U 1 ) = cosi? = 1 (U 1 )
dr(U 2 ) = sin# = 2 (U 2 )
dr(U 3 ) = = 3 (U 3 ).
Since the 1 forms ^ and dr are equal on the frame field Uj they are
equal on all tangent vectors and vector fields (by Lemma 1.5.4), and
so 0j = dr.
Similarly 2 (U) = ILE 2 , so
02(Ui) = sin i? 2 (U 2 ) = cos t$ d 2 (U 3 ) = 0.
Also
dd(U i ) = U i [d] =— .
OX;
i> = tan" 1 ] —
x
64
so
M334 II.8
a#_ l ,y
y
r sin#
r 2
sini?
r
3x y\ 2 x 2
x 2 + y 2
3.5_ 1 Jl\
x
r cos #
r 2
cos #
r
ay 1+ y*W
X 2 + y 2
3z
Thus . a
r sin tf
rdt?(U 1 ) = —=62(1],)
r cos 1? ^ ,
r d#(U 2 ) = = 2 (U 2 )
r d#(U 3 ) = = 2 (U 3 )
and so 2 = r d#.
3 (Ui) = 3 (U 2 ) = 3 (U 3 ) = 1
dz(Ui) = dz(U 2 ) = dz(U 3 ) = 1
and so 3 = dz.
(b) Ejr] =dr(E i ) = 1 (E i ) = 5 li .
2 (Ei) 5 2 i
Ei [tf] =dt?(Ei)=^ = T .
E^z] = dz(Ei) = 3 (Ei) = 5 3i .
(c) Ejf] = df(Ei) = I — dr + — d# + — dz (Ej) by the theorem
to be proved in the final section,
3f . 1 df . 3f
= 5„ +  _ 6 2 i + _ 5 3 i.
3r r ov dz
Notice how we have used the Kronecker delta in (b) and (c) to
convey several pieces of information in one statement. It does not
matter if your solution did not use it, but bear in mind that you can
often shorten the amount of writing by using it.
Now we check the structural equations.
X = dr, 2 = rd#, 3 = dz,
co 12 = d#, co 13 = 0, co 23 =0 as found in Section II.7:
d0! = d(dr) = O and w 12 a 2 + co 13 a 3 = d& a rd# + = 0;
d0 2 = d(rd#) = dr a d# and co 21 a B x + co 23 a 3 = dd a dr + = dr a d#;
d0 3 = d(dz) = and co 31 aM co 32 a 2 = 0;
dco 12 = d(d#) = and co 13 a co 32 = 0;
dco 13 = and C0i 2 a co 23 = 0;
dco 23 = and co 21 a co 13 = 0.
M334 II.8
3. Page 95, Exercise 1.
d0 = d(2 fj0i)
65
= 2 d(f^i)
i
= 2 (dfi a 0. + fjd^j),
i
= 2 (d^ A fl. + f; 2 CO" a 0.),
i J
= 2 dfj a 0j + 2 2 qcoy a 0j,
= 2 {df: + 2 !>••} A 0,
j i J J
by Theorem 1.6.4.(2),
by Theorem 11.8.3.(1),
by rearrangement,
66
M334 II
PARTIAL DIFFERENTIATION
We have met two new coordinate systems in this chapter: r, d, z and p, d, <£. If you
have met these before you may have come across expressions such as 9f/9r, 9f/9#
and equations such as
9f 9f 9f
df = — dp + — d# + — dip.
dp 9# 9<p
Since the only partial derivatives we have so far formally defined are the 9f/3x,
where x are the natural coordinate functions, our purpose here is to give a meaning
to expressions such as 9f/9r, 9f/9# which is consistent with our idea of a partial
derivative.
To do this properly, we must first of all be quite clear about the use of the symbols
x, y, z, as defined on page 4. They denote not "dummy variables", but the natural
coordinate function from E 3 to R. That is, if p = (p 2 , p 2 , P3 ) then x(p) = pi , y(p) =
P2> z (p) = P3 In particular (x, y, z)(p) = (pj , p 2 , p 3 ) = p, and so (x, y, z) is the
identity function on E 3 .
Now suppose that f is a function from E 3 to R. We must first notice that, although f
is a single function from E 3 to R, it has different expressions in terms of x, y, z and
in terms of p, #, <p say. For example, let us consider the function f = x 2 + y 2 + z 2 . We
know p 2 = x 2 + y 2 + z 2 , so we can also write f = p 2 . Now, because (x, y, z) is the
identity function, we can write
f = f o ( x , y, z) = f(x, y, z)
but f is not equal to f(p, #, <p) because
f(p, 0, <p) = f o (p, 0, J) = p 2 + # 2 + <p 2 .
Can we find a function g such that f = g(p, #, <p), that is f = gom where m is the
function (p, #, <p): E'
E 3 ?
E ;
m= (p, #,</>)
In this case we can do so quite simply. If p € E 3 then
f(p) = P 2 (P) = (P(P)) 2
g (p, T>, ^)(p) = g(p( P ), 0(p), spi$))
and so we require the function g to pick out the first coordinate and square it: but
the function that does this is precisely the function we have called x 2 . Thus g = x 2
and so
f =x 2 ° m.
We return to the general case, and recall that the definition of 9f/9x is
9f d
T(P) = j f (t> P* Ps)
ox dt
, = ^7 f(t, y(p), 4p))
t=pj dt
t=x(p).
M334 II 67
Roughly speaking, we are holding y and z fixed and differentiating with respect to x.
If we want to define 3f/3p in a similar fashion we need to hold # and <p fixed and
differentiate with respect to p. This is where we need to use the function g: since
f = g(p, &, if)
3f
we define — by
dp
3f d
^(p) = dT g(t, ^p),^(p))
t=p(p)
But this is precisely 3g/3x evaluated at (p(p), #(p), <^(p)), because, roughly speaking,
3g/3x means "differentiate with respect to the first variable" no matter what the
"variables" may be called.
This is formalized in the following definition.
Definition
If m = (m lf m 2 , m 3 ) is a differentiable onetoone function from a subset of E 3 to
E 3 with differentiable inverse m" 1 , and f is a differentiable function from E 3 to R,
and g is a function such that f = g o m (in fact g must be f ° m" 1 ), then
3f 3g
— (p) = £ (m(p)).
dm ox
Notice that this, definition of 3f/3mj is dependent on the whole function m: if m =
(in! , m 2 , m 3 ) then 8f/3m^ may be different depending on whether m or m is being
used.
In practice this definition is very easy to work with. In our previous example we had
f = P 2 , g = x 2 , and so 3f/3p = (3g/3x)(m)= 2x(p, &, \p) = 2p. Another example: if
f = p cos if then g = x cos z, and
3f = 3g
(m) = cos z(p, &, <p) = cos ip;
op ox
3f 3 g/
— =(m =0;
3t? 3y V }
3f 3g. .
— ■— (m) =xsinz (p, r?, <^)= psimp.
o<p oz
This should show you that this process effectively treats p, #, <p as "dummy variables"
just as x, y, z are normally treated, and so the new partial derivatives can be written
down without the step of introducing g.
For the usual partial derivatives 3f/3x, 3f/3y, 3f/3z we had the result (Corollary
1.5.5):
df =— dx + — dy + — dz.
ox 3y 3z
Now that we have defined other partial derivatives suitably, we can prove an
analogous theorem.
68 M334 II
Theorem
If m = (mi , m 2 , m 3 ) is a differentiable onetoone function from a subset of E 3 to
E 3 with differentiable inverse m" 1 , and f is a differentiable function from E 3 to R,
then
3f
df = 2— dm.
bm i 1
In particular,
df dr+  d# + dz;
3r ou oz
3f 3f in 3f J
df = —dp +  d0 + — d*.
3p 3# 3<p
(Corollary 1.5.5 is the particular case of this theorem for mj = Xj.)
Proof
We use the chain rule given in the additional text on Section 1.7. We can take
g=f°m' 1 to obtain
f = g o m.
Now the chainrule gives
If v is a tangent vector at the point p, this gives
U v ) = g *( m (p)) m *( v )>
(*)
where g is written as a matrix evaluated at m(p) and the vectors f^v), m^v) are
written as column vectors. Now, Theorem 1.7.5 tells us that
f # (v) = (v[f]) = (df(v)):
this is a 1 X 1 matrix so that corresponding column vector is also (df(v)). The same
theorem tells us that
m *( v ) = ( v t m i] > v t m 2l » v t m 3] )
= (dm^v), dm 2 (v), dm 3 (v)),
which is
'dmi(v)
dm 2 (v)
as a column vector.
By Corollary 1.7.7 g^has matrix
'9g 9g_ dg \
k 9xi' 3x 2 ' 3x 3 
M334 II
Evaluating this at the point m(p) we have
\
Ji (m(p)), b ± (m(p)), *J (m(p))
OX! dx 2 ox 3
69
which is precisely
3f 3f 3f
om l 3m 2 am 3 '
by the definition of (df/8m). Putting the expressions we have obtained into (*):
(df(v)) =
af af af
x~ (p)> r— (p), r — (p)
onij dm 2 om 3
dm^v)
dm 2 (v)
dm 3 (v)
9f 3f 3f
— (p)dm,(v) + — (p)dm 2 (v) + _ (p)dm 3 (v)
om! 3m 2 3m 3 v '
3f 9f 3f \ \
= — dmj + — dm 2 +  — dm 3 (v)
\om l dm 2 om 3 
This is true for all tangent vectors v, and so
3f 3f 3f
df = dm» + — dm 2 + dm 3 .
m! om 2 om 3
You do not need to know any details of this proof. All you need for this course is
the defintion of general partial derivatives and the result stated in the theorem.
Summary
Notation
[elf 3f 3fl
l3r' 3#' 3zJ
(df df 3fl
3p'3#' dd
Text, page 66
Text, page 66
Definition
(i) Partial derivatives
Text, page 67
Results
af 3f 3f
(I) df=~dr + — d#+ — dz.
3r 3# 3z
af 3f 3f
(II) df = dp+— d^+~d«^.
op 3# dip
Text, page 68
Text, page 68
70
Techniques
(i) Finding partial derivatives.
(ii) Using the expressions for df in terms of the
partial derivatives.
M334 II
Text, page 67
Text, page 68
Exercises
Technique (i)
3f 3f df
1. Find — , — , — when f = p costf cos </?.
dp 3# by
df 3f 3f , r . ,
2. Find — ; — , — when f = x 4 + ir.
3r 30 3z
Technique (ii)
This has already been tested in Exercise 11.8.4(c). Do that exercise now if you
previously deferred it.
Solutions
1. f = pcos# COS<£
= x cos y cos z (p, #, if)
= g(p, #, <p) where g = x cos y cos z.
— = — (p, #, \p) = cosy cosz (p, #, <p) = cost? cos<p.
3p 3x
— = — (p, #, tp) = x siny cos z (p, d, <p) =  p sin# cos (p.
3# 3y
— = — (p, #, <p) =x cosy sinz (p, #, </?) =  p cos# sin<p.
3<p 3z
2. We must first of all express f in terms of r, &, z.
From the diagram we see that
x = r cos &
M334II 71
and so
f=x 2 + z 2
= r 2 cos 2 & + z 2 .
As before, we can introduce g = x 2 cos 2 y + z 2 , or we can go directly to the
results:
9f
— = 2r cos 2 i?,
3r
3f
— = 2r*cosi? sin#;
3f
A" =2z 
oz
72 M334 II
FURTHER EXERCISES AND SOLUTIONS
Section II. 1
Recommended Exercises
Page 49, Exercise 2. This shows that Euclidean distance is a metric in the sense of
M202andM231.
Page 50, Exercise 11. This relates the vector operation grad to the norm and dot
product.
Solutions
2(a). d(p, q) = p  q = (2(p  qj) 2 ) 2 , which is nonnegative.
d(p, q) = => 2(pj  qj) 2 = 0, and a sum of squares is zero if and only if each term
is zero, that is pj = qj.
2(b). ( Pi  qj) 2 = (qj  Pi ) 2 so d(p, q) = d(q, p).
2(c). d(p, q) + d(q, r) = p  q + q  r > p  q + q  r = jp  r = d(p, r).
11(a). df(v) = 2 Vi _l = (Sarins i Ui(p)) = vVf(p).
3xy 3xi
11(b). Putw= (Vf)(p).
u[f] = uVf(p) = u«w
and luwl < INI w = w
with equality when u = 5^ .
w
Section II. 2
Technique Exercise
Page 55, Exercise 1.
Other Recommended Exercises
Page 55, Exercise 5. This deal with unitspeed repararmetrizations in the sense of
Theorem II. 2.1.
Page 56, Exercise 10. This shows that arclength is preserved by reparametrizations
which are symmetric in the sense of Comment (i).
Solutions
1(a). (2,2^, (2, 2, 1) (2) 1; ,
t 2 + 2,3;
<°' 2 > 2t W (0 ' 2 ' 2) (2,U)
M334 II 73
1(b). s(t) = 2t + ;p
5. If /3j is the unitspeed reparametrization based at t = tj, and sj is the arc
length function based at t = tj,
«(t)=^i(s 1 (t))=^ 2 (s 2 (t)).
Si(t) = j v(u)du
ft 2 ft
= J v(u)du + I v(u)du
= s + s 2 (t)
where s is the arclength of a from t t to t 2 .
10. Put0 = a(h).
(a) Arclength of from a to b
b
J v 0( u ) du
V(d)
= J h'(u)v a (h(u))du
h'(c)
= J v a (t)dt
c
= arclength of a from c to d.
Arclength of 3 from a to b
= J V£(u)du
a
fh _1 (c)
= J (h'(u))v a (h(u))du
h" 1 (d)
= J v a (t)dt
d
= f v^tjdt.
c
74 M334 II
Section II.3
Technique Exercises
Page 6364, Exercises 2, 3.
Other Recommended Exercises
Page 64, Exercise 5. This expresses all the Frerret formulas in a single format.
Page 64, Exercise 6. This deals with the approximation of a curve by part of a circle.
Solutions
(l+s)*(ls)* 1
2. T(s) =
N(s) =
B(s) =1
2 V2
{ 22T (H> a ,o
\ V* ' V2 f
(1+S)2 (1S) 5 1
T ' T 'V2
k(s) = r(s) = _^___ i
W V2(ls 2 ) 2
3. Evaluate both sides of each equation.
5. AXT = kBXT = kN
A X N = tT X N + kB X N = tB  kT
AX B = rTX B=rN.
6. The only solution for 7 is
s s
7(s) = c  r cos  N + r sin —T
r r
Nn
where c = — + j3(0)
/c
1
r =
Section II.4
Technique Exercises
Pages 7475, Exercises 1(a), 3(a), 4 (first part), 6, 7, 9(a), (b).
Other Recommended Exercises
Page 74, Exercise 5. This gives a formula for K.
Page 75, Exercise 13. This introduces another special curve derived from a given
curve. Since one of a, a is negative, you will first have to adapt Example II.3.3 for
the case where a is negative.
M334 II 75
Solutions
1
2+t 2
1
!( a )' T(t)=— 2 (2,2t,t a )
N(t) = — 2 (2t,2t 2 ,2t)
B(t)=^L(t 2 ,2t,2)
2
fc(t) = r(t) =
3(a). T(0) =
(2+t 2 ) 2 *
(1,0,1)
V2
N(0) = (0, 1, 0)
B(0) = tl^il
K(0) = 1
3
T(0)=.
4. Evaluate both sides of each equation.
6. a' = cT.
a" = cT' = c 2 /cN.
a'x a" = c 3 /cB.
Now substitute in Theorem II.4.3.
7. T(t)=^sint,^cost,)
N(t) = (cost, sint, 0)
, lb b a\
B(t) =  sin t, — cos t, —
\c c c/
«(t) =i 2
r(t)=K
9( a). B JL2d>
V2
#=
4
7(t) = ( t  tVt+ ).
9(b). u = (0,0,1)
* = *
4
7(t) = (3t  t 3 , 3t 2 , 0).
76 M334II
dv _
5. a" = — T + kv 2 N
dt
ll«" 2 =[~ 2 +(/CV 2 ) 2 .
3. S S
13. Ifa>0, then fc(s) = =, N(s) = ( cos, sin, 0);
c c c
if a < 0, then k(s) = — «, N(s) = (cos, sin, 0).
c c c
Substitute these to obtain
0ab* = "ah
for all a. Now (a) = a so
<%b* ='<W
Section II. 5
Technique Exercises
Page 80, Exercises 1(b), 2(b), (d), (e), (f), 4.
Other Recommended Exercises
Page 80, Exercise 3, This extends the result that if Y is a vector field on a curve and
11Y is constant then YY' = 0.
Page 80, Exercise 6(b). This deals with the definition of covariant derivative.
Page 80, Exercise 7. This exercise introduces the bracket of two vector fields.
Solutions
1(b). (l,2,4) p
2(b). yU 3 .
2(d). sinx U! + cosx U 3 .
2(e). y 2 cosx \J X  y 2 sin x U 2 .
2(f). (y 2  xcosx yzsinx)U! + (xsinx +yzcosx)U 2  2xy U 3 .
4. X = Sxj Uj.
V v X = 2 V [ Xi ] Uj = S 2 v: ^ Ui = S 2 Vj 5 ij Ui
v . i i . j a Xj i j J J
ZviUV.
W(p+tv)W(p+tv) = constant
=»W(p+tv)W(p+tv)' =
=>W(p)V v W = 0.
7(a). Both sides equal X 2
i J
dw: dv
v j — " w j —
3 Xj dx
M334 II 77
6(b). W(a)'(t) = V a , (t) W = V v W.
* 3f
Bx:
J / 1
7(b). The definition is antisymmetric.
7(c). [U,[V,W] ] [f] = U[V,W] [f]  [V,W]U[f]
= UVW[f]  UWV[f]  VWU[f] +WVU[f].
Adding the other two similar expressions to this gives zero.
7(d). [fV,gW] [h] = fV[gW[h] ]  g W[fV[h] ]
= f(V[g]W[h] +gVW[h]) g(W[f]V[h] + fWV[h])
= fV[g] W[h]  gW[f] V[h] + fg[V,W] [h].
Section II. 6
Techinque Exercises
Pages 8485, Exercises 3, 4.
Solutions
3. Choose any unit E 2 orthogonal to E 1 and put E 3 = Ej X E 2 . For example
E 2 = sin z U 2  cos z U 3
E 3 =  sin x Uj + cos x cos z U 2 + cos x sin z U 3 .
4. The same formulas are obtained as for the spherical frame field.
Section II. 7
Technique Exercises
Pages 9091, Exercises 1, 2, 3, 7.
Solutions
df
1. <^i20;o; 13 = co 23 =  7 .
2. All zero.
3. A is orthogonal.
oj 12 = df;
co 13 = cosf df;
co 23 = sinf df.
78
7(a). For any f,
M334 II
Fifl =cosiz> cos v ^— + cos kd smj? —  + sin (/>».
11 J ^ ox r 3v oz
8y
p = (x 2 + y 2 + z 2 ) 5
t^tan 1 y
U/
<p = tan
l
^(x 2 +y 2 ) 2 '
evaluate the partial derivatives and substitute.
7(b).  sin p F 2 + cos p F 3 .
Section II.8
No further exercises are recommended.
M334 II 79
DIFFERENTIAL GEOMETRY
I Calculus on Euclidean Space
II Frame Fields
III Euclidean Geometry
IV Calculus on a Surface
V Shape Operators
VI Geometry of Surfaces in E
4 PART III THE OPEN UNIVERSITY
— 9
is* '<■>■,*■:.;■■ ■
51" • Mathematics: A Third Level Course
DIP
PART III EUCLIDEAN GEOMETRY
DIFFERENTIAL GEOMETRY
9
THE OPEN UNIVERSITY
Mathematics: A Third Level Course
DIFFERENTIAL GEOMETRY PART III
EUCLIDEAN GEOMETRY
A commentary on Chapter III of O'Neill's
Elementary Differential Geometry
Prepared by the Course Team
THE OPEN UNIVERSITY PRESS
Course Team
M334 III
Chairman:
Mr. P.E.D. Strain
Dr. R.A. Bailey
Lecturer in Mathematics
Course Assistant in Mathematics
With assistance from:
Dr. J.M.
Mr. G.J.
Dr. P.M.
Mr. P.B.
Dr. F.C.
Mr. T.C.
Mr. RJ.
Dr. C.A.
Mr. M.G.
Aldous
Burt
Clark
Cox
Holroyd
Lister
Margolis
Rowley
, Simpson
Senior Lecturer in Mathematics
Lecturer in Educational Technology
Lecturer in Physics
Student Computing Service
Lecturer in Mathematics
Staff Tutor in Mathematics
Staff Tutor in Mathematics
Course Assistant in Mathematics
Course Assistant in Mathematics
Consultants:
Prof. S. Robertson
Prof. T. Willmore
Professor of Pure Mathematics,
University of Southampton
Professor of Pure Mathematics,
University of Durham
The Open University Press, Walton Hall, Milton Keynes, MK7 6AA
First publishedV^S^eprinted 1979
Copyright © 1975 The Open University
All rights reserved. No part of this work may be reproduced in any form, by mimeograph or any other means,
without permission in writing from the publishers.
Produced in Great Britain/by
The Open University Pros
ISBN 335 05702 0>y
This text forms part of the correspondence element of an Open University Third Level Course. The complete list
of parts in the course is given at the end of this text.
For general availability of supporting material referred to in this text, please write to: Open University
Educational Enterprises Limited, 12 Cofferidge Close, Stony Stratford, Milton Keynes, MK11 1BY, Great
Britain.
Further information on Open University courses, may be obtained from The Admissions Office, The Open
University, P.O. Box 48, Milton Keynes, MK7 6AB.
1.2
M334 III
CONTENTS Page
Set Book 4
Bibliography 4
Conventions 4
III. 1 Introduction and Isometries of E 3 5
III. 2 The Derivative Map of an Isometry 14
III. 3 Orientation 20
III. 4 Euclidean Geometry 26
III. 5 Congruence of Curves 32
Further Exercises and Solutions 38
4 M334 III
Set Book
Barrett O'Neill: Elementary Differential Geometry (Academic Press, 1966). It is
essential to have this book: the course is based on it and will not make sense without
it.
Bibliography
The set books for M201, M231 and MST 282 are referred to occasionally; they are
useful but not essential. They are:
D.L. Kreider, R.G. Kuller, D.R. Ostberg and F.W. Perkins: An Introduction to
Linear Analysis (Addison Wesley, 1966).
E.D. Nering: Linear Algebra and Matrix Theory (John Wiley, 1970).
M. Spivak: Calculus, paperback edition, (W.A. Banjamin/AddisonWesley 1973).
R.C. Smith and P. Smith: Mechanics, SI edition (John Wiley, 1972).
Conventions
Before starting work on this text, please read M334 Part Zero. Consult the Errata
List and the Stop Press and make any necessary alterations for this chapter in the set
book.
Unreferenced pages and sections denote the set book. Otherwise
O'Neill denotes the set book;
Text denotes the correspondence text;
KKOP denotes An Introduction to Linear Analysis by D.L. Kreider, R.G.
Kuller, D.R. Ostberg and F.W. Perkins;
Nering denotes Linear Algebra and Matrix Theory by E.D. Nering;
Spivak denotes Calculus by M. Spivak;
Smith denotes Mechanics by R.C. Smith and P. Smith.
References to Open University Courses in Mathematics take the form:
Unit Ml 00 22, Linear Algebra I
Unit MST 281 10, Taylor Approximation
Unit M201 16, Euclidean Spaces I: Inner Products
Unit M231 2, Functions and Graphs
Unit MST 282 1, Some Basic Tools.
M334 III.l 5
HI.1 INTRODUCTION AND ISOMETRIES OF E 3
Introduction
The aim of this chapter is to prove that a unitspeed curve in E 3 is determined, apart
from its position in space, by its curvature and torsion functions, which were intro
duced in Section II. 3; in other words, that the curvature and torsion of a curve, as
functions of the arclength, tell us all there really is to know about it. We cannot
begin to prove this until we have decided what we mean by "apart from position in
space": when are two curves in different positions in E 3 going to be declared
"essentially the same"? We shall not finally answer this question until Section III.5,
but we make a start in this section by defining isometries, that is mappings of E 3
which preserve distance. You have met some isometries before — translations in Unit
M201 15, Affine Geometry and Convex Cones, Section 1.1, and orthogonal linear
transformations in Unit M201 24, Orthogonal and Symmetric Transformations,
Section 1. You should remind yourself of the results of the latter section now, as
they are used in this section. Apart from that, this section depends only on Sections
1.7 and II. 1; the later sections of Chapter II are not referred to yet.
READ: Introduction to Chapter III and Section III.l (pages 98 102).
Comments
(i) Isometries An important point about the isometries of E 3 is that they
form a group; that is, that they satisfy the following conditions.
(a) The composition of two isometries is again an isometry — this is
Lemma 1.3.
(b) Composition of isometries is associative — this is true for com
position of any functions.
(c) There is an identity isometry I such that IF = FI = F for all iso
metries F — this is true because the identity mapping I is an
isometry.
(d) Each isometry F has an inverse F" 1 which is also an isometry. This
condition requires slightly more checking. It is an immediate con
sequence of Definition 1.1 that isometries are onetoone mappings,
for if F(p) = F(q) then
d(p, q) = d(F(p), F(q)) = d(F(p), F(p)) =
and so p = q. The fact that isometries of E 3 are also onto mappings
follows from Theorem 1.7, because the translation T is onto and the
orthogonal transformation C is onto, so their composite F = TC must
also be onto. Since F is onetoone and onto F has an inverse func
tion F" 1 = C" 1 !" 1 , which is a mapping because both C" 1 and T" 1 are
mappings (see Comments (ii) and (ill)). Now
d ( F " 1 (p)»F 1 (q)) = d(FF' 1 (p), FF _1 (q)) because F is an
isometry
= d(p, q) because FF" 1 = I
and so F" 1 is indeed an isometry.
6 M334 III.l
(ii) Translations The translations form an important set of isometries, as
noted in Example 1.2(1). We shall use the notation T a to denote translation
by a, that is,
Then
T a(p) = P + a.
Tt>T a (p) = p + a + b = T a + b (p) = T b + a (p)
so T b T a is translation by b + a. In fact the translations correspond to the
points of E 3 in such a way that the composition of translations corresponds
to ordinary vector addition; hence the commutativity of composition of
translations follows from the commutativity of vector addition. We have the
following commutative diagram:
a, b
corresponding
translation
T a> T b
■► a + b
composition
corresponding
translation
► T a T b = T a + b
Since the identity mapping I is translation through 0, T a = T_ a and we
have this commutative diagram also:
additive inverse
corresponding
translation
corresponding
translation
(T a r = T_ a
Thus the translations also form a group.
(iii) Orthogonal Transformations As noted in Lemma 1.5, these form
another important set of isometries. The definition given on page 100 is in
fact equivalent to that given in Unit M201 24, Orthogonal and Symmetric
Transformations, page 8, because all the spaces in which we are interested
are finitedimensional. Thus we can make use of the result proved in that
unit, that the following conditions on a linear transformation C of finite
dimensional Euclidean spaces are all equivalent:
(a) C preserves dot products (that is, C is an orthogonal transformation);
(b) The rows of the matrix C are orthonormal;
M334 III.1 7
(c) The columns of the matrix C are orthonormal;
(d) t CC = C*C = I; that is, C" 1 = *C (that is, C is an orthogonal matrix);
(e) C maps some orthonormal basis to another orthonormal basis;
(f) C maps all orthonormal bases to orthonormal bases.
Since, for orthogonal transformations Q, Cj, C,
(CiC 2 ) = C 2 C 1 = C 2 Ci = (CiCj)
and
\cr l ) = Yc) = C = (C 1 )' 1
this immediately tells us that the composite of two orthogonal
transformations is an orthogonal transformation and that the inverse of an
orthogonal transformation is an orthogonal transformation. Thus these also
form a group.
(iv) Page 101: Theorem 1. 7 This breakdown of an isometry F as TC is very
useful. In future work with isometries we shall nearly always use this ex
pression, sometimes without explicitly defining T and C. If F t = TiCi and
F 2 = T 2 C 2 are isometries then their composite F t F 2 is also an isometry
(Lemma 1.3) so we can write F 1 F 2 = T 3 C 3 , where T 3 is the translation part of
FxF 2 and C 3 is the orthogonal part of FxF 2 . Then
T3C3 = TiCiT 2 C 2
but it is not generally true that T 3 = TjT 2 (although C 3 = CiC 2 ) because Ci
and T 2 do not commute in general. (See the Theory Exercises if you are
interested in investigating this further.)
Supplementary Comments
(i) Page 99: rotation The formulas for the coordinates of q in terms of
those of p were worked out in Unit M201 24, Orthogonal and Symmetric
Transformations, Section 1.2.
(ii) Page 101: lines 10 and 9 If we put r = ap + bq, then rj = apj + bqj for
i = 1, 2, 3. Using the identity established in line  11,
F(r) = 2riF(ui), F(p) = 2 Pi F(ui), F(q) = 2 qi F(ui)
so
F(r) = 2r i F(u i )
= 2 (a Pi + b qi )F(ui)
= 2ap i F(u i ) + 2bq i F(u i )
= aZp i F(u i )+bSq i F(ui)
= aF(p) + bF(q).
8
Summary
M334 III.l
Notation
F
T
T
C
I
Definitions
(i) Isometry F
(ii) Translation by a, T a
(iii) Orthogonal transformation C
(iv) Identity mapping I
Example
(i) Rotation about the z axis
Results
(i)
(«)
The composite of two isometries is again an
isometry.
For translations:
(iii)
(iv)
(v)
(vi)
(vii)
T a T b = T b T a = T a + b>
(T a r 1= T a .
Given p, q in E 3 , there is a unique translation T
such that T(p) = q.
If T is a translation and T(p) = p for any p, then
T = I.
Translations are isometries.
Orthogonal transformations are isometries.
Any isometry fixing is an orthogonal trans
formation.
(viii) Each isometry F has a unique expression as TC,
where T is a translation and C is an orthogonal
transformation. Moreover T = Tjr(o).
(ix) Every isometry has an inverse isometry.
Techniques
(i) Recognition of isometries, translations, orthogonal
transformations.
(ii) Expression of an isometry in standard form TC.
(iii) Calculation of the image of a point under an iso
metry expressed in standard form.
Page 98, Definition 1.1
Page 98, Example 1.2(1)
Text, page 6
Page 100, line 11
Page 99, line 6
Page 98, Definition 1.1
Page 98, Example 1.2(1)
Page 100, line 10
Page 99, line 6
Page 99, Example 1.2(2)
Page 99, Lemma 1.3
Page 99, Lemma 1.4(1)
Page 100, Lemma 1.4(2)
Page 101, Lemma 1.4(3)
Page 100, lines 89
Page 98, Example 1.2(1)
Page 100, Lemma 1.5
Page 100, Lemma 1.6
Page 101, Theorem 1.7
Text, page 5.
Page 98, Definition 1.1,
Page 98, Example 1.2(1), and
Page 100, line 10.
Page 101, Theorem 1.7
Page 102, line 5
M334 III.l
Exercises
Technique (i)
1. Page 103, Exercise 4.
2. Page 103, Exercise 6, (a) — (d), first part.
Technique (ii)
3. Page 103, Exercise 6, second part.
Technique (Hi)
4. Page 103, Exercise 5(a).
Theory Exercises (omit if short of time)
5. Page 103, Exercise 1.
6. Page 103, Exercise 2.
7. Page 103, Exercise 3.
Solutions
1. Page 103, Exercise 4.
As noted in Comment (iii), C is orthogonal <=► C X C = I •*= > the rows of C are
orthonormal. We check the last condition as it involves only six calculations.
.11. lUl 2 _J.Ll (4 + 4 + 1)=1
2. 1
3' 3'
3' 3'
 }, " 114(4 + 1 + 4) = !
\F' 3' 3J \3' 3' 3/ 9 (1 4 4) l
2 2
3' 3'
2 1
3' 3'
Ll (  4 + 2 + 2) =
HflMl)=i (2 + 2  4) = ()
This shows that C is orthogonal.
C(p) =
v
2
3
19
3
]_
10
M334 III.l
2.
C(q) =
lKR)(l)=i(107649)=iP = 15.
pq = 3.1 + 1.0 + (6)f3) = 3  18 =  15.
C( P )C(q) (§)( 4) + ji„ „., 3)l3)
Page 103, Exercise 6.
In each case we must check whether d(F(p), F(q)) = d(p, q) for all p and q.
(a) F(p)=p.
d(F(p),F(q)) = d(p,q)
= p(q)
= (l)(pq)
= I" 1 I IIP" q (using ap = a p)
= ipq
= d(p, q).
This is true for all p and q, so F is an isometry.
In this case F: pi ypaa is just the perpendicular projection of E 3
(b)
onto the onedimensional subspace spanned by a. We can easily see
from the picture
that F does not preserve norms: for example, we can choose p to be
orthogonal to a and then F(p) = so F(p) ¥= p. Since F does
not preserve norms, F cannot preserve dot products.
Alternatively we proceed as follows.
d(F(p), F(q)) = d(pa a, qa a)
= paa qaa
= (pa qa)a
= p*a q*a a, as above,
= p*a q*a, since a = 1,
= l(pqH
= lpqINIcost?,
where t? is the angle between
pq and a,
= d(p, q) cosi?
since Mai
= 1.
which is not equal to d(p, q) unless # is a multiple of it.
M334III.1 11
As yet another approach, since the image of F is not the whole of
E 3 , F is not onto E 3 and so F cannot be an isometry, by Comment
(i).
(c) F(p) = (p 3  l, P2 2, Pl 3).
d(F(p), F(q)) = d((p 3  1, p a " 2, Pl  3), (q 3  1, q 2 " 2, qi  3))
= (P3 l,p 2 2,p 1 3)(q 3  l,q 2 2, qi 3)
= (P3 q3»Pa q 2> Pi qi)
= ((P3q 3 ) 2 + (Paq 2 ) 2 + (piqi) 2 ) 1
= (pi  qi» P2  qa, p 3  q3)
= lpql
= d(p, q).
This is true for all'p and q, and so F is an isometry.
(d) F(p) = ( Pl ,p 2 ,l).
d(F(p), F(q)) = d(( Pl , p 2 , 1), (q lf q 2 , 1))
= (Pi> Pa> 1) " (qi, qa, 1)
= (Pi qi.Pa" q2, 0)
= ((Piqi) 2 + (P2qa) 2 ) 21 ,
which is not equal to d(p, q) unless p 3 = q 3 . For example,
d((0, 0, 0), (0, 0, 1)) = 1 but
d(F(0, 0, 0), F(0, 0, 1)) = d((0, 0, 1), (0, 0, 1)) = 0.
Thus F is not an isometry.
Again, we could note that, since the image of F is
{ p G E 3 : p 3 = 1 } =£ E 3 ,
F is not onto, and so cannot be an isometry.
3. Page 103, Exercise 6.
(a) F(p) =_ p and so F = I, which is an orthogonal transformation.
This is already in standard form, and so T is the identity, C = I.
(b) F is not an isometry.
(c) F(0) = (1, 2, 3) so T, the translation part of F, is translation by
(1,2,3). If F = TC,
F(p)=TC(p)
i.e. (p 3  l,p 2  2, Pl  3) = C(p) + (l,2,3)
i.e. (p 3 , p 2 , Pl ) + (1, 2,3) = C(p) + (1, 2, 3)
so C(p) = (p 3 , p 2 , pj)
/0 1^
so C has matrix 10
10
(In fact this represents reflection in the plane x = z.)
(d) F is not an isometry.
12
4.
M334 III.l
5.
Page 103, Exercise 5(a).
F(p) = T a C(p)=a + C(p)
/ 1
V2
1
V2
/
V2
10
V2.
1  V2
1 + V2,
Page 103, Exercise 1.
Put F = CT a . Since C and T a are isometries F is an isometry (by Lemma
III.l. 3), and thus by Theorem III.l. 7 we can write F = T^D where T. is
translation through b and D is an orthogonal transformation. Now
CT a (p) = T b D(p) for all p
i.e.
i.e.
C(a + p) = b + D(p)
C(a) + C(p) = b + D(p).
Putting p = gives C(a) = b because C(0) = D(0) = 0.
Substituting back, we have
i.e.
and so
Thus
b + C(p) = b + D(p)
C(p) = D(p)
C = D.
F = T b D = T C ( a )C.
for all p
for all p
M334 III.1
13
Alternatively, we could show that CT a and T£/ a \C have the same effect on
every point of E 3 . For p in E 3 ,
CT a (p) = C(a + p) = C(a) + C(p),
while
T C ( a )C(p) = T C(a) (C(p)) = C(a) + C(p).
6. Page 103, Exercise 2.
F = T a A and G = T b B, so
FG = T a AT b B.
By the result of Exercise III. 1.1,
AT b = T A(b) A
and so
FG = T a AT b B = T a T A(b) AB.
T a T A(b) is the translation through a + A(b); AB is an orthogonal trans
formation, being the product of two orthogonal transformations. Thus FG
has translation part T a + 'A(b) anc ^ orthogonal part AB.
Similarly GF has translation part T b + jj/ a \ and orthogonal part BA.
7. Page 103, Exercise 3.
F = T a C. Since T a and C both have inverses we know that
F" 1 = (TaC)' 1 = C' 1 ^ 1 .
Now
C^V 1 = C _1 T_ a , by Lemma 111.1.4(2),
= Tci(_ a )C _1 , by Exercise III. 1.1,
= T^i/^C" 1 , by linearity of C" 1 .
Thus the translation part of F" 1 is T_c;i( a ); the orthogonal part of F _1 is C" 1 .
14
M334 III.2
III.2 THE DERIVATIVE MAP OF AN ISOMETRY
Introduction
This section follows on from Sections III.l and 1.7. We also need to use certain
results from M201 — that there exists a unique linear transformation taking one
basis of E 3 to another basis of E 3 (see Unit M201 2, Linear Transformations,
Section 1.8) and that a linear transformation is orthogonal if and only if it takes an
orthonormal basis to another orthonormal basis (see Unit M201 24, Orthogonal and
Symmetric Transformations, Section 1.2).
The purpose of this short section is to show that there is a unique isometry taking
any given frame in E 3 to any other given frame in E 3 . In order to do this we have to
look at the derivative map F*: we find that this takes a particularly simple form
when F is an isometry, for if F = TC then F* is represented simply by C.
READ: Section III. 2 (pages 104 106).
Comment
(i) Page 105: lines 18 to 15
F^p and C cannot be the same function, for
they act on different spaces — T p (E?Ji andE 3 respectively. However, T p (E 3 ) is
isomorphic to E 3 via the "canonical" isomorphism D : E 3 ► T D (E3) given
by P P
»V T
(v G E 3 ).
3»
To say that F^ and C "differ only by the canonical isomorphisms of E
means, in effect, that the following diagram commutes; that is, that F +p and
C send vectors that correspond to each other via 0« to vectors that corres
pond to each other via <b„, x .
F(P)
* T F(P) (E3 >
► F *p( v p) = Cv
F(P)
Supplementary Comment
(i) Page 105, line 4 If v = (v 1} v 2 , v 3 ) then the ith coordinate of Cv is 2q;vj,
so that J
(Cv) =22c i jVjU i (q)
for any point q. In particular, putting q = F(p) gives
(CV) F(P) = ?fijvjUi(F(p)).
M334 III.2 15
But F *( v p ) = ( Cv )f( p )' and so
^(v p ) = SZcyVjUifFfc)) = 2 c ijVj Ui(F(p)).
1 J 1 »J
F,
However, v p = 2)v;Uj(p), so this becomes
fJx
SvjUjCp)! = i 2c i jv j U i (F(p)),
which can be abbreviated to the result given in line 4 by omitting the "p"s
and writing Ui for Uj(F).
Summary
Results
(i) The derivative map of an isometry F = TC sends
the tangent vector v at the point p to the tangent
vector Cv at the point F(p). Page 104, Theorem 2.1
(ii) If F is an isometry, F* p is an orthogonal trans
formation from T (E 3 ) to Tp, ,(E 3 ). Page 105, Corollary 2.2
(iii) Given two frames in E 3 , there is a unique isometry
taking one onto the other. Page 105, Theorem 2.3
Techniques
(i) Finding the derivative map of an isometry using
Result (i). Page 104, Theorem 2.1
(ii) Finding the isometry mentioned in Result (iii),
using the method employed in the proof of
Theorem 2.3. Page 105, line 3 of text
Exercises
Technique (i)
1. Page 106, Exercise 1.
Technique (ii)
2. Page 107, Exercise 5.
Theory Exercises (omit if short of time)
3. Page 106, Exercise 2. Write F = T a A and G = T^B and use the results of
Exercise III. 1.2, whose solution occurs on Text page 13, and Exercise
III. 1.3, whose solution occurs on Text page 13.
4. Page 107, Exercise 4(a). (HINT: r is in the plane through p orthogonal to q if
and only if (r  p)«q = 0.)
16
Solutions
M334 III.2
Page 106, Exercise 1.
We can write T = TI for any isometry T, where I is the identity mapping. In
the case when T is itself a translation this expresses T in standard form, as I
is an orthogonal transformation. Using Theorem III.2.1 we have
T *( v ) = ( Iv )t( P )
'T(p)'
because Iv = v.
Thus T*(v) has the same Euclidean coordinates as v; only the point of
application has changed. In other words, T se (v) is parallel to v.
Page 107, Exercise 5.
The attitude matrix of the e frame is
u
A =
and the attitude matrix of the f frame is
/ 1
B =
V2
1
V2
By the argument at the end of the text on page 106 we must have that
/ 1
C = X BA =
V2
V2
1_
V2
2
3

V2
_J_
V2
72"
2
3
V2 V2 V2,
W
M334 III.2
and T is the translation through
17
qc( P ) =
/
l
_2
3
V2
2
3
^ \372 372 V2/\°/
i
w
=
4
"3
W
Thus the isometry F is given by
/M/*
1
V2
F(r) =
4
3
+
2
3
1
3
2
3
1
1^ 2 J
1
4
1
\3y/2 V2 V2^
for all points r = (r 1} r^ r 3 ) of E 3 .
Let us check that F really does carry the e frame into the f frame; that is,
F^ejJ = fj for i = 1, 2, 3;that is, F(p) = q and Cej = fj for i = 1, 2, 3.
F(p) =
1
2v/2
fr
V2
2
3
1
3
2
3
1
4
1
\V2
V2
V2
18
M334 III.2
/• \ I
1
2>/2
I »\
\V2/
1
\ '
Ce! =
V2
2
3
1
V2
2
3
V2 V2 V2
1
V2
Ce,=
V2 V2 V2
Ce,=
V2
_2
3
V2 V2 V2
Page 106, Exercise 2.
If F = T a A and G = T^B, where A, B are the orthogonal parts of F and G,
then, by the result of Exercise III. 1.2, GF has orthogonal part BA. Hence, by
Theorem III.2.1, if v p is any tangent vector,
(GFMv p ) = (BAv) GF(p) .
The same theorem gives
F*(v p ) = Av F(p)
and, applied to the isometry G and the tangent vector F^v),
G*(F*(v p )) = B(Av) G(F(p)) ,
M334 III.2 19
that is
(G*Fj(v p ) = (BAv) GF(p) .
Hence (GF)*(v p ) = (G^F^Vp) for all tangent vectors v p , and so it
follows that (GFJ* = G^F*.
By the result of Exercise III.1.3, F" 1 has orthogonal part A" 1 , and so
Theorem III. 2.1 gives
(F')»(v p )  (Av) Fl(p)
for any tangent vector v p . We must show that this is the same as (F*)" 1 (v p ).
We know that
F*(v p ) = ( Av ) F(p) >
that is, F* acts on the vector part of v« by the orthogonal transformation A
and on the point of application of v p by F. Thus (F*)" 1 , which is the inverse
of F*, must act on the vector part of v p by A" 1 and on the point of
application of v p by F" 1 ; that is,
(FJ 1 ^) = (AM F i (p) 
Consequently
(F*) 1 ^) = (F%(v p )
for all v p , and so (F*) 1 ^ 1 )*.
4. Page 107, Exercise 4(a).
r is in the plane through p orthogonal to q if and only if
(rp).q = 0.
C preserves dot products so this happens if and only if
(C(r P )).C(q) = 0.
Suppose T is translation by the vector a.
C is linear, so
C(rp) = C(r)C(p)
= C(r) + a(C(p) + a),
= F(r)F( P ).
Thus
(C(rp)).C(q) = (F(r)F(p)).C(q),
which is zero if and only if F(r) is in the plane through F(p) orthogonal to
C(q). Hence F carries the plane through p orthogonal to q to the plane
through F(p) orthogonal to C(q).
20
M334 III.3
III.3 ORIENTATION
Introduction
This section continues from Section III. 2, also tying in ideas from Unit M201 24,
Orthogonal and Symmetric Transformations, Sections 2.2 to 2.4. You may recall
from that unit that frames in E 3 are of two distinct types, known as lefthanded and
righthanded, distinguished by which hand "best fits" the frames. Moreover, ortho
gonal transformations of E 3 also fall into two distinct sets: there are the proper
rotations (that is, those transformations physically realizable in 3space as a rotation
about an axis) and the improper rotations (that is, those transformations which are
not realizable as an actual movement in 3space because they consist of a proper
rotation followed by a reflection). It turns out that proper rotations preserve the
"handedness" of frames, whereas improper rotations interchange lefthanded and
righthanded frames, and that the corresponding matrices have determinant +1 for a
proper rotation and  1 for an improper rotation.
In this section we give a more mathematical definition of "righthanded" and "left
handed" and classify all isometries of E 3 (not just orthogonal transformations) in
terms of their effect upon frames. We are able to state and prove the results quite
precisely, in a form that makes them suitable for use in proving the important
theorems of Section III.5.
READ: Sections III. 3 (pages 107110).
Comment
(i) Page 107: Remark 3.1
(1) In this case the attitude matrix is I, and det I = +1.
(2) Since e 3 is a unit vector orthogonal to both ej and e 2 , it must be
either e!Xe 2 or (e^ej). By the result of Exercise 11.1.4(d),
ei*e 2 X e 3 = e 3 '(e 1 Xe 2 ), which is +1 if e 3 = e! X e 2 , 1 if e 3 = ~(c\ X e 2 ).
(3) This is a restatement of part of Lemma II. 1. A, on Text, Part II,
page 7.
Supplementary Comments
(i) Page 110: the determinant Remember that a formal expansion of
yields
ei
e 2
e 3
Vl
v 2
v 3
w x
w 2
w 3
(v 2 w 3  v 3 w 2 )e! + (v 3 W!  V!W 3 )e 2 + (vjw 2  v 2 Wi)e 3 .
M334 III.3 21
(ii) Page 110: proof of Lemma 3.5 If the frame is negatively oriented, the
ex component of v X w is
v 2 e 2 X w 3 e 3 + v 3 e 3 X w 2 e 2 = (v 2 w 3  v 3 w 2 )e 2 X e 3
= (v 2 w 3  v 3 w 2 )(e!), by Remark 3.1(3),
= e (v 2 w 3  v 3 w 2 )e!
because e =  1 when the frame is negatively oriented.
(iii) Page 110: end of proof of Theorem 3.6
F*(v) X F*(w) = e
ei
e 2
«s
Vl
v 2
v 3
Wj
w 2
w 3
= e ((v 2 w 3  v 3 w 2 )e! + . . .)
= e ((v 2 w 3  v^F^Uifr)) + • • •)
= e (F 5lc ((v 2 w 3  v^U^p) + . . .)),
= e
/
\
U^p) U 2 (p) U 3 (p)
vi v 2 v 3
Wi w 2 w 3
1
= e F^v X w),
and
e = ei .e 2 X e 3 = F :)c (U 1 (p)).F 5!c (U 2 (p)) X F*(U 3 (p))
= (sgnF)U 1 (p).U 2 (p)XU 3 (p),
by Lemma III.3.5,
because F% is
linear,
by the definition
of cross product,
by Lemma III.3.2,
= sgnF,
because the frame
U!( P ), U 2 (p),
U 3 (p) is posi
tively oriented.
Summary
Notation
sgnF
Page 108, line 8
Definitions
(i)
(")
Positively oriented )
Righthanded f
Negatively oriented \
Lefthanded J
Page 107, line 17
Page 107, line 16
22 M334 III.3
(iii) Sign of an isometry, sgn F Page 108, line 6
(iv) Orientationpreserving Page 109, Definition 3.3
(v) Orientationreversing Page 109, Definition 3.3
Results
(i) The natural frame field is positively oriented. Page 107, Remark 3.1(1)
(ii) The frame e 1? e 2 , e 3 is positively oriented if and
only if e 2 X e 2 = e 3 . Page 107, Remark 3.1(2)
(iii) Frenet frames are positively oriented. Page 107, Remark 3.1(2)
(iv) If e u e 2 , e 3 is a positively oriented frame,
e[ = e; X e^ =  e^ X e;
for (i, j, k) = (1, 2, 3) or (2, 3, 1) or (3, 1, 2). Page 107, Remark 3.1(3)
(v) F^(e 1 )F ! ,(e 2 ) X F*(e 3 ) = sgn F e r e 2 X e 3 . Page 108, Lemma 3.2
(vi) The coordinates of tangent vectors in terms of a
positively oriented frame give the expected for
mula for the cross product; if the frame is nega
tively oriented the usual formula must be multi
plied by  1. Page 110, Lemma 3.5
(vii) Orientationpreserving isometries preserve cross
products; orientationreversing isometries change
their sign:
F^v X w) = sgn F F^v) X F*(w). Page 110, Theorem 3.6
Techniques
(i) Distinguishing positively and negatively oriented
frames. Page 107, lines 17, 16
(ii) Finding the sign of an isometry Page 108, lines 6 to 8
(iii) Finding the effect of an isometry on a cross pro
duct, using Result (vii). Page 110, Theorem 3.6
Exercises
Technique (i)
1. Find the orientation of each of the frames in Exercise III. 2. 5 on page 107.
Technique (ii)
2. Find the sign of the isometry in Exercise III.2.5 and check that Lemma
III.3.2 is satisfied.
Technique (iii)
3. Page 111, Exercise 3.
M334 III.3
Theory Exercises
23
(omit if short of time)
4. Page 111, Exercise 1. (HINT: Use the result of Exercise III. 1.2.)
5. Page 111, Exercise 2.
Solutions
1. The attitude matrix of the e frame is
2 2 1
3 3 3
111
3 3 3
\l _2 2 /
\3 3 3/
whose determinant is
2l.2_2j_2\\ + 2/2a_(_42\ + l/_2/ 2\ 1.1 1
3\3 3 3 \ 3J] 3\3 3 \ 3/ 3/ 3\ 3 \ 3/ 3 3
= ^(2(2 + 4) + 2(2 + 4) +(41))
= 1,
so the e frame is positively oriented. Alternatively (and more simply)
e,Xe 2 = ±(2.2  1.1, 1.2  2.2, 2.1  2.(2))
= I(3,6,6) = e 3
and so the e frame is positively oriented by Remark 3.1(2). Similarly,
fi* h =72(0.0 1.1, 1.0 1.0, 1.1 0.0)
V2
(l,0,l)=f 3
and so the f frame is negatively oriented.
Lemma III.3.2 is satisfied if F 5N (e 1 )»F Jc (e 2 ) X F*(e 3 ) = sgn F e^ X e 3 . We
found in the previous section that
1
C =
1
V2
2
3
V2 V2 '372j
V2
_2
3
24 M334 III.3
sgn F . det c = _>_(. _ _^ . 1. _y + + j 2 [lfc  Lfe
=^ei88i)=i.
Since sgn F =  1, Lemma III.3.2 predicts that the e frame and the F(e) frame
(that is the f frame) have opposite orientations: we have already shown that
this is the case in the previous exercise.
3. Page 111, Exercise 3.
We want to show that C*(v X w) = sgn C C*(v) X C^w).
v = (3, 1,1)
w = (3,3, 1)
so
v X w = (1  3, 3  3, 9 + 3) = (2, 0, 6).
C is its own orthogonal part, so C* is represented by the same matrix as C.
C*(v) =
4 i
C*(w) =
1
3
11
3
7_
"3
C*(vX w)=
C*(v) X C*(w) = ( 1, 3, 1) X i ^, ]
lf\
8_
3
14
3
, 10 8 14, n . v .
so the formula is checked if sgn C =  1.
M334 III.3 25
However,
2 1 _ 2 2\_ l/2.2 _ 1.1
3 3 3 3 3 3 3 3 3
sgnC = detC=[ + )+f(
= ^(2(2 + 4) + 2(24)(4l))
= 1
and so the formula is checked.
4. Page 111, Exercise 1.
Let F = T a A, G = T^B. Then, by the result of Exercise III.1.2, FG has ortho
gonal part AB. Hence
sgn (FG) = det (AB), by definition,
= (det A) (det B)
= (sgn F)(sgn G), by definition of sgn F, sgn G.
Similarly
sgn (GF) = (sgn G)(sgn F)
= (sgn F)(sgn G).
Putting G = F" 1 , gives
sgn (FF" 1 ) = (sgn F)(sgn F' 1 ).
But FF" 1 = I, the identity mapping, and so
(sgn F)(sgn F" 1 ) = sgn I = +1 :
therefore
1 1
sgn F" 1 =
sgnF
= sgn F, because sgn F = ±1.
Page 111, Exercise 2.
Let G be an orientationreversing isometry of E 3 . We want to express G as
HqF, so we define F to be Ho _1 G. Then certainly
H F = H Hq^G = G
and so all that remains to be shown is that F is orientationpreserving. Now,
by the result of the previous exercise,
sgn G = sgn(H F) = (sgn H )(sgn F).
Here G and Hq are both orientationreversing, and so
sgnG = sgnH = l;
hence sgn F = +1 and so F is orientationpreserving.
26 M334 III.4
III.4 EUCLIDEAN GEOMETRY
Introduction
This section depends on Sections III. 1 III. 3 and Section II. 3. Now we are able to
use the ideas and results of Sections III. 1  III. 3 to begin to define what we mean by
"the same, but for position in space, as . . .". Loosely speaking, we are going to
declare two "objects" in E 3 (for example, curves or vector fields) to be "the same
but for position in space" if there is an isometry carrying one into the other. In
order that this definition agree with our intuitive ideas of "sameness", two con
ditions must hold: (a) the relation "is the same, but for position in space as" must
be an equivalence relation (any sensible definition of "same" must lead to an
equivalence relation); (b) any properties of the objects concerned that we regard as
essential, such as the curvature of a curve, the derivative of a vector field, must be
preserved by isometries, so that objects which are "the same" have the same
essential properties. Condition (a) is easily checked: the relation is transitive because
the composite of isometries is itself an isometry (Lemma III. 1.3); it is reflexive
because I is an isometry; it is symmetric because each isometry has an inverse iso
metry (Comment (i) on Section III. I). Condition (b) is the concern of this section,
where it is shown that isometries preserve derivatives of vector fields on curves and
the Frenet apparatus of unitspeed curves.
READ: Section III.4 (pages 112115).
Additional Text
(i) ArbitrarySpeed Curves and Isometries While it is true that any mapping
F preserves the velocity of a curve a in the sense that
F^cx') = (F(o0)'
(Theorem 1.7.8), the speed of a curve is not generally preserved, because
a' and (F(a))' are not necessarily the same. However, in the case where
F is an isometry an argument similar to that given at the beginning of the
proof of Theorem 4.2 shows that F does preserve speed: F preserves norms,
as noted on page 105, and so
v F(o) = l(F(«))'li
= F.(«')H
= Hall
= V
a*
We have shown that a and F(a ) have the same speed if F is an isometry. The
following exercise shows that they have the same curvature, and plusor
minus the same torsion.
Let a be an arbitraryspeed regular curve in E 3 , F an isometry of E 3 ,
= F(a), and a a unitspeed reparametrization of a, where a = a(s).
Prove that K Q = K a and T Q = sgn F r a . (HINT: Use the above result to
prove that j3  F(a) is a unitspeed reparametrization of j3 in the sense of
Theorem II. 2.1, and then use Theorem III.4.2.)
M334 III.4 27
Supplementary Comments
(i) Page 113: second half As usual we are writing F = TC and C = (qj).
(ii) Page 114: fourth line of proof T and T are defined by T = 0',
T = j3 ', so the result follows using the fact that 3 ' = (F(j3))' = F*(j3') by
Theorem 1.7.8.
(iii) Page 114: seventh line of proof K and k are defined by K = jS",
K=p"; j3"=F*(0") by Corollary HI.4.1; F*(j3") = 0" \\ because
F % preserves norms.
(iv) Page 114: eleventh line of proof (¥^"))/k = F*(j3"/k) because F^
is a linear transformation at each point.
Summary
Results
(i) Isometries preserve derivatives of vector fields on
curves: (F*(Y))' = F*(Y').
In particular, isometries preserve acceleration. Page 113, Corollary 4.1
(ii) Isometries preserve the speed of curves. Text, Page 26
(iii) Isometries preserve the Frenet apparatus of unit
speed curves, except that orientationreversing iso
metries change the sign of the torsion and reverse
the direction of the binormal:
k = k T=F !tc (T)
r = sgnFT N=F*(N)
B = sgn F F !C (B) Page 114, Theorem 4.2
(iv) A similar result for arbitraryspeed curves: if
j8 = F(a), then Kq = K a and r Q = sgn F r a . Text, Page 26
Techniques
(i) Applying Result (i) to find the derivatives of vector
fields of the form F 5C (Y) on a curve F(a). Page 113, Corollary 4.1
(ii) Using Result (iii) to find the Frenet apparatus of
curves of the form F(a). Page 114, Theorem 4.2
28
Exercises
M334 III.4
Technique (i)
2. Page 115, Exercise 2. (Remember that, because C is orthogonal, C* has the
same matrix as C.)
Technique (ii)
3. Page 115, Exercise 1.
Solutions
We have seen that a and 3 have the same speed, and consequently a. and /3
have the same arclength function s. Now
?(s) = (F(ff))(s) = F(n(s)) = F(a) = j8
and so j3 is a unitspeed reparametrization of )3 in the sense of Theorem
II.2.1.
Hence
by the definition of the curvature and torsion of an arbitraryspeed curve.
Similarly
K a =K S (s) r a =7ff(s).
However, ZX is unitspeed and j8 = F(S) and so Theorem III.4.2. gives
r^=sgnFr s .
K.Q K s
Hence
and
K P = K P^ = K ^ = Kql
T 8 = T /^ S ) = Sgn F T ^ S ^ = Sgn F T «*
Page 115, Exercise 2.
 1 cost
a(t) = C(cx(t)) =
\/2 V2
1
V2
V2
sin t
2t
cost
^(sint 2t)
,^(sint + 2t)
/
M334 III.4
29
Y(t) = C*(Y(t)): since C is itself an orthogonal transformation C* has the
same matrix as C and so
l
C*(Y(t)) =
\
" V2 "72
\° 72 72/
1t 2
\ 1+t2 /
V2t 2
V2
That is,
Thus
Now
and so
Y(t) = (t,V2tV2).
Y'(t) = (l,2V2t,0).
Y'(t) = (l,2t,2t)
t
C*(Y'(t)) =
1 1_
72 "V2
1 1
\/2 V2
2t
2t
2V2t
Thus Y'(t) = C*(Y'(t)) for all t, and so Y' = C*(Y').
a'(t) = (sint, cost, 2),
a"(t) = (cos t, sin t, 0)
30
so
/
C*(a"(t)) =
1
V2
1
V2
1
V2
1
V2
/ cost 1
sint
cost
V2
1
V2
suit
sint
whereas
M334 III.4
a'(t) = sint,i(cost  2), ^(cost + 2)
— «/,v I . sint sint'
a(t) = cost, ^ ^
and so C*(a"(t)) =a"(t) for all t; that is, C*(a") = 2e".
We have already computed Y', a ", Y , a": from this we see that
(Y'V)(t)= (l,"2t, 2t)(cost, sint, 0)
= cost + 2tsint
and
so
(Y'.«")(t) = ( 1, 2V2t, 0) cos t,
= cost + 2t sint
Y •« = Y a .
sint sint
V2' V2
3. Page 115, Exercise 1.
(a) Since 3 has unit speed, T = ]3', and so if /3 is a cylindrical helix then there is a
constant unit vector u and a constant angle # such j3'«u = cos #. Let
ii = F^u). Then
H» = f,<u)H = Hi = i
and so u is a constant unit vector. Moreover, F(j3) also has unit speed; that is,
T = F(8)', and so
Tu = (F(0))'.u
= (F(0))'.F»
= F*(/3>F*(u),
= COS #,
and so F(j3) is also a cylindrical helix.
by Theorem 1.7.8,
because F ^preserves dot products,
M334III.4 31
(b) Let F(j3) denote the spherical image of F(j3), T the unit tangent vector field
on F(j3). Then F(p) is the curve whose points have coordinates given by T,
while ft is the curve whose points have coordinates given by T. In other
words
T(t)=3(t)0 (t) and T(t) = F$)(t) F (/?(t))
Omitting the t:
T = J?0 and T=F03) F(i3 ).
By Theorem III.4.2, T = F^(T)
andsoF(jS)F^) = F Hc (33).
Since F = TC, Theorem III. 2.1 tells us that
F*(00) = CflJ ) F 0)
and so
^)F(P) = C(?) F( ^
and so
F(0) = C(/3).
32 M334 III.5
ffl.5 CONGRUENCE OF CURVES
Introduction
At last we reach the definition of congruence of curves promised in Section III.l
This is the definition suggested in Section III. 4 and, as the remarks there show,
congruence is in fact an equivalence relation. Results from all the earlier sections of
Chapter III are used to prove the main theorem that unitspeed curves are congruent
if and only if their curvatures are the same and their torsions differ at most in sign.
Using the definitions of Section II.4 we can widen this theorem to include arbitrary
speed regular curves. Finally we generalize the main theorem to nonFrenet frame
fields, using the ideas of Section II.6. This last theorem (Theorem III.5.7) is less
important now but will be needed in Chapter VI: you are not expected to remember
its details.
Some of the proofs in this section are a little long and technical. Try to work
through them, but if you get bogged down make sure that you at least understand
the statements of the theorems: they are the whole purpose of this chapter.
READ: Section III.5 (pages 116 121).
Comments
(i) Pages 117118: proof of Theorem 5.3 Notice the structure of the
proof. If F is to carry a to j3 then certainly F(a(t)) = 3(t) for all t, so our first
requirement is that F(a (0)) = 0(0). Secondly, we know from Theorem
III.4.2 that F* must take the Frenet frame of a at a (0) to the Frenet of j3 at
0(0), reversing the direction of the binormal if r a =r^, so we require that
F* have this effect on T a (0), N a (0), B a (0). These two conditions specify F
uniquely, so the final stage is to check that F(a ) is actually equal to j3.
Notice also the beauty and economy of the last stage of the proof. We know
that T«T<1 because
TT = fflTcost>,
where # is the angle between T and T, and cos # = 1 if and only if # is a
multiple of 2ir, that is if and only if T and T are equal. Thus we can show
that T = T by showing that TT is always 1. However, because the same
remarks apply to N and N, B and B, we can show that all three pairs are
equal by the single step of showing that
f = TT + NN + BB= 3.
This is shown by a technique we have noted before, that of proving f' = to
show f is constant and noting the f(0) has the correct value.
(ii) Page 119: Corollary 5.5 When O'Neill uses the word "helix"
unqualified, he means not a general cylindrical helix, but a circular helix,
that is a cylindrical helix whose crosssection curve is (part of) a circle (see
Exercises II.4.8 and II.4.10), as he remarks in Example 1.4.2(2). An equiva
lent definition of circular helix is as a curve congruent to a reparametrization
of the helix
1 1 — ► (a cos t, a sin t, bt)
given in Example 1.4.2(2), and it is this definition O'Neill uses in the proof
here.
M334 III.5 33
Supplementary Comments
(i) Page 117: line 9 F is orientationpreserving because it carries one
Frenet frame into another, and Frenet frames are always positively oriented.
(ii) Page 117: line 2
k = k. w by Theorem III.4.2,
= Kg, by the hypotheses of the theorem.
(iii) Page 119: Example 5.4 The "a" used here is simply a shorthand for
l/\/2 = 1/c: it is not the same "a" as in Example II.3.3.
(iv) Page 120: first two lines of proof a and 3 must be based at the same
value of t so that a. and 3 have the same arclength function s, for if 6t is
based at tj and ]8 at t 2 then
s a (0 = \ v a ( u ) du
J U
= \ Vg(u)du, because v a = v„,
= \ V u ^ du+ \ V u ^ du
= s + S/3 (t),
and the constant s~ is zero if and only if t x = t 2 .
(v) Page 121: Theorem 5. 7, hypotheses (1) and (2) These are just natural
extensions of the hypotheses in Corollary 5.6, for if {Ej} and {Fj} are the
Frenet frame fields on a, j3 respectively then condition (1) becomes
simply v a = v~ and condition (2) becomes K a = k„, T a = t«.
(vi) Page 121: lines 7; 8 of proof Since F carries the Ej frame at a (0) to the
Fj frame at j3(0), the two lefthand equations are straightforward. For the
other two:
a'«Ej = (F(a ))'«F !( .(Ej), by definition,
= F^a )«F^(Ej), because F,,. preserves velocities,
= a Ej, because F^ preserves dot products,
= j3'*Fj, by hypothesis (1);
also
E}Ej = (F 5lc (E i ))'.F :JC (Ej), by definition,
= F 5C (E)'F % (E), because F* preserves derivatives of
vector fields,
= Ej «Ej, because .F* preserves dot products,
= Fj'«Fj, by hypothesis (2).
34 M334 III.5
(vii) Page 121: line 9 of proof Since {Ej} is a frame, Ej' has an orthonormal
expansion Ej = 2 ajjEj, where aj; = Ej'«Ej (Theorem II. 1.5). Similarly,
Fj' = 2 bjjFj, where by = Fj'Fj. We have just shown that Fj'Fj = Ej'*Ej> so
D ij ^ a ij and we can write Fj' = 2 ajjF;.
(viii) Page 121: lines 10, 11 of proof Ej«Ej = 5jj =^(Ej'Ej)' =
==>Ej'.Ej + Ej.Ej' = 0=
•a ij + aji = 0.
(ix) Page 121: line 13 of proof
f = 2(Ej'.F i + Ej.Fj')
i
= 2Ej'.Fj + SE;.Fj'
j j j j
= .2(aijEjFj + a ji E j .F i )
1 > J
= 2(a i j + ajj)Ej.F i .
lj J
(x) Page i2i: /me i4 of proof For i = 1, 2, 3, since_Ej(t) and Fj(t) are unit
vectors, Ej(t)Fj(t) < 1 with equality if and only if Ej(t) and Fj(t) are equal.
Thus f(t) = 2 Ej(t)Fj(t) < 3 with equality if and only if each Ej(t) is equal
to the corresponding Fj(t). We have shown that f' is the zero function, so f
itself is a constant function. However, f(0) = 3 (because Ej(0) = Fj(0) for
each i) and so f(t) = 3 for all t, that is Ej(t) is equal to Fj(t) for all t.
Summary
Definitions
(i) Congruent curves
(ii) Parallel curves
Page 116, Definition 5.1
Page 116, line 1
Results
(i)
If a '(s) and j3'(s) are parallel for each s, a and j3 are
parallel. Page 117, Lemma 5.2
(ii) If a and j3 are parallel and a(s Q ) = 0(s o ) for some
s^, then a = ]3.
(iii) Unitspeed curves a, 3 are congruent if and only if
K a = Kp and T a = ±Tp
(iv) Arbitraryspeed regular curves a, ]3 are congruent if
and only if
Page 117, Lemma 5.2
Page 117, Theorem 5.3 and
Page 114, Theorem 4.2
Page 120, Corollary 5.6 and
Text, page 26
M334 IH.5 35
(v) If a, 3 are arbitrary curves and {Ej}, {Fj} are
frame fields on each such that
a'Ei = j3'Fi
and Ej'«E; = Fj'F;,
then a and j3 are congruent. Page 121, Theorem 5.7
Techniques
(i) Recognition of congruent curves. Page 116, Definition 5.1
(ii) Finding an isometry establishing the congruence of
two curves. Page 117, proof of
Theorem 5.3
Exercises
Techniques (i) and (ii)
1. Page 122, Exercise 6.
Theory Exercise
2. Page 122, Exercise 5.
Solutions
1. Page 122, Exercise 6.
a(t) = (v/2t, t 2 , 0)
a'(t) = (v/2, 2t, 0)
v«(t) = V2(l + 21 2 ) 2 "
a"(t) = (0, 2, 0)
We use the formulas of Theorem II.4.3.
a'(t) X a"(t) = (0, 0, 2V2)
so
so
2>/2 1
Ka(t)_ 2 v / 2(l + 2t 2 f (l + 2tf
a'"(t) = so r a (t) = 0.
3(t) = (t,t,t 2 )
/3'(t) = (l, 1, 2t)
v^(t) = (2 + 4t 2 ) i =V2(l + 2t 2 ) i = v a (t)
0"(t) = (0, 0, 2)
0'(t) X 0"(t) = (2, 2, 0)
2y/2 1
K {t) ' 2V2(1 + 2t 2 )f " (1 + 2t 2 )T " Ka(t) 
/3"'(t) = so T0(t) = O = r a (t).
36 M334 HI.5
Thus <x and /3 have the same speed, curvature and torsion and so they are
congruent.
a'(0) = (v/2, 0, 0) so T a (0) = (1, 0, 0);
a'(0) X a"(0) = (0, 0, 2^2) so B a (0) = (0, 0, 1);
N a (0) = B a (0) X T a (0) = (0, 1, 0):
0'(O) = (1,1,0) so Tp(0) = l^,±0\;
0'(O) X /3"(0) = (2, 2, 0) so B0(O) =^,^,o);
N^(0) = B^O) X T^(0) = (0,0,1).
Since r a = ra we want to find the isometry F carrying the a Frenet frame to
the j3 Frenet frame: such an isometry has orthogonal part with matrix
C =
1
V2
1
V2
1
V2
V2
V2
1
V2
1
V2
1
V2
transpose of attitude
matrix of 3 frame
attitude matrix of (X
frame
Since a(0) =]3(0) = 0, C(a(0)) = 8(0) and so the translation part of the F is
just the identity.
Checking:
1
72
72
F(a(t)) =
1
V2
1
V2
1
=«t).
Alternatively, since t ol = tq also, we could choose F to take the a Frenet
frame to the frame Tj3(0), N/j(0), B«(0). This gives
C =
1
V2
1
72
1
V2
1
01 [72 ° 72
72
1
72
and again the translation part of F is the identity.
M334 III.5 37
2. Page 122, Exercise 5.
If a and are congruent, then K a = kq = k and T a = ± T0 = r. Suppose F
is an isometry such that F(a) = )3. Then, as noted in the Additional Text (i),
we must have
F(«(0)) = «0)
F !e (T a (0)) = T^(0)
F*(N a (0)) = N^(0)
F*(B a (0)) = sgn F 1^(0) (*)
by Theorem III.4.2. This theorem also tells us that
T3 = sgn F r a .
If r # 0, this equation tells us that sgn F = 1 if T a = Tg, sgn F =  1 if
r a = tq. Thus the four equations (*) determine the isometry F uniquely,
and the proof of Theorem III.5.3 shows that in fact F(oc) = j3.
If r = then there is no constraint on sgn F, so we can have sgn F = 1 or
sgn F = 1. In each of these two cases the equations (*) give a unique iso
metry F. Moreover, since r a = tr and T a =  tq, the proof of
Theorem III.5.3 shows that both of these isometries satisfy F(a) = 3.
38 M334 III
FURTHER EXERCISES AND SOLUTIONS
Section III.l
Technique Exercises
Page 103, Exercises 5(b), (c).
Solutions
5(b). (V2,5,V2)
5(c). ( V2, 1, V2).
Section III. 2
Technique Exercise
Page 107, Exercise 4(b). (HINT: (0, 1, 0) has the same length as (1, 0,1)^2.)
Other Recommended Exercise
Page 106, Exercise 3. This proves the result given at the end of page 106 in an
alternative way.
Solutions
4(b). There are many possible solutions, each of which is obtained by putting a
particular value of ft in the following.
sin ft 1 cos ft
V2 72" "72"
C = cos ft sin ft
sin ft 1 cos ft /
72" V2 x/2 /
and T is translation through
i j. sm ^ + _1 o _ cos # i + sin ft 1_
1 2^2 >/2 2 ,x 2V2 V2
3(a). The translation part of F carries C(0) = to p, so is Tp.
The orthogonal part of F carries Uj(0) to c v so has the coordinates of the ej
as its columns, so it is A.
M334 in 39
3(b). The eframe at p is taken to the natural frame at by
(TpA^ATp^TA^A.
The natural frame at is taken to the fframe at q by TqB" 1 .
(TqB _1 )(TA(p)A) takes the eframe at p to the fframe at q and has ortho
gonal part B"*A.
Section III.3
Recommended Exercises
Page 111, Exercise 4. This shows that orthogonal transformations with determinant
+1 can be realized as rotations of E 3 .
Page 111, Exercise 6. Some more group theory.
Page 111, Exercise 7. This looks at isometries of E 1 and E 2 , which are defined in
the obvious way.
Solutions
4. Put e 3 = (p/p). Choose e! and e 2 to be orthogonal unit vectors spanning
the subspace of E 3 orthogonal to e 3 . C(ex) is also orthogonal to e 3 and has
unit length, so
C(ei) = cos & e! + sin d e 2
for some #. If
C(e 2 ) = sin i? e!  cos # e 2 ,
replace & and e 2 by their negatives.
6. For both parts: det (AB) = det A det B
det A 1 = (det A) 1
detl = +l.
7. E 1 :F(x) = ex + a
f cos & e sin ^
E 2 : F = TC, C =
\
 sin ^ e cos #
\
In each case, orientationpreserving if e = +1.
Section III.4
Technique Exercise
Page 116, Exercise 3.
40
Solution
3.
M334 III
T(3,3)
Section III.5
Technique Exercises
Page 122, Exercises 2,4. (HINT for Exercise 2: The Frenet apparatus is not needed
— look for an isometry first.)
Other Recommended Exercises
Page 121, Exercise 1. This gives another characterization of congruence of curves.
Page 122, Exercise 3. This uses Theorem III.5.7 and the connection forms.
Page 122, Exercise 11. This proves a theorem equivalent to Theorem III.4.2 and
Theorem III. 5. 3 for curves in the plane.
M334 III
Solutions
41
2. There are eight possible solutions. Each of e, 6, 17 may be ±1 in the follow
ing.
7 (t) = (eV2t,53t 2 ,V2t 3 )
4.
72
F =
*
V2
JL.
K=ir=i
a=2,b = 2.
/..
V3
2
1
2
F =
1
1.
1
2
V3
IT
1.
3.
11.
If F = T p C, where C(uj) = ej, then F(a) = 0.
The frame field on a is (Ej(a)}, that on 5 is {Ej(0)}.
a'.Ei(a)=0i( a ').
(Ei(a))'.Ej(a) = (V a 'Ei).Ej(a) = «ij(a').
If: let F be the isometry taking
a(0) to 0(0)
T a (O)toT0(O)
N a (0) to eN0(O) where kq = e/c a .
Follow the proof of Theorem III.5.3.
Only if: follow the proof of Theorem III.4.2.
DIFFERENTIAL GEOMETRY
I Calculus on Euclidean Space
II Frame Fields
III Euclidean Geometry
IV Calculus on a Surface
V Shape Operators/
VI Geometry of Surfaces in E 3
M33' PART IV
516.
I
DIF
THE OPEN UNIVERSITY
Mathematics: A Third Level Course
9
PART IV CALCULUS ON A SURFACE
DIFFERENTIAL GEOMETRY
9
THE OPEN UNIVERSITY
Mathematics: A Third Level Course
DIFFERENTIAL GEOMETRY PART IV
CALCULUS ON A SURFACE
A commentary on Chapter IV of O'Neill's
Elementary Differential Geometry
Prepared by the Course Team
THE OPEN UNIVERSITY PRESS
Course Team
M334 IV
Chairman:
Mr. P.E.D. Strain
Dr. R.A. Bailey
Lecturer in Mathematics
Course Assistant in Mathematics
With assistance from:
Dr.J.M. Aldous
Mr. GJ. Burt
Dr. P.M. Clark
Mr. P.B. Cox
Dr. F.C. Holroyd
Mr. T.C. Lister
Mr. R.J. Margolis
Dr. C.A. Rowley
Mr. M.G. Simpson
Senior Lecturer in Mathematics
Lecturer in Educational Technology
Lecturer in Physics
Student Computing Service
Lecturer in Mathematics
Staff Tutor in Mathematics
Staff Tutor in Mathematics
Course Assistant in Mathematics
Course Assistant in Mathematics
Consultants:
Prof. S.Robertson
Prof. T. Willmore
Professor of Pure Mathematics,
University of Southampton
Professor of Pure Mathematics,
University of Durham
The Open University Press, Walton Hall, Milton Keynes
First published ^76vkeprinted 1980
Copyright © 1976 The Open University
All rights reserved. No part of this work may be reproduced in any form, by mimeograph or any other means,
without permission in writing from the publishers.
Produced in Great Britain By
The Open University Pre/s
ISBN 335 05703 9
This text forms part of the correspondence element of an Open University Third Level Course. The complete list
of parts in the course is given at the end of this text.
For general availability of supporting material referred to in this text, please write to: Open University
Educational Enterprises Limited, 12 Cofferidge Close, Stony Stratford, Milton Keynes, MK11 1BY, Great
Britain.
Further information on Open University courses may be obtained from The Admissions Office, The Open
University, P.O. Box 48, Milton Keynes, MK7 6AB.
1.2
M334 IV 3
CONTENTS
Page
Set Book 4
Bibliography 4
Conventions 4
Introduction 5
IV.l Surfaces in E 3 6
IV.2 Patch Computations 20
IV. 3 Differentiable Functions and Tangent Vectors 33
IV.4 Differential Forms on a Surface 43
IV.5 Mappings of Surfaces 49
IV. 7 Orientability 56
Further Exercises and Solutions 59
4 M334 IV
Set Book
Barrett O'Neill: Elementary Differential Geometry (Academic Press, 1966). It is
essential to have this book: the course is based on it and will not make sense without
it.
Bibliography
I
The set books for M201, M231 and MST 282 are referred to occasionally; they are
useful but not essential. They are:
D.L. Kreider, R.G. Kuller, D.R. Ostberg and F.W. Perkins: An Introduction to
Linear Analysis (Addison Wesley, 1966).
E.D. Nering: Linear Algebra and Matrix Theory (John Wiley, 1970).
M. Spivak: Calculus, paperback edition (W.A. Benjamin/ AddisonWesley, 1973).
R.C. Smith and P. Smith: Mechanics, SI Edition (John Wiley, 1972).
Conventions
Before starting work on this text, please read M334 Part Zero. Consult the Errata
List and the Stop Press and make any necessary alterations for this chapter in the set
book.
Unreferenced pages and sections denote the set book. Otherwise
O'Neill denotes the set book;
Text denotes the correspondence text;
KKOP denotes An Introduction to Linear Analysis by D.L. Kreider, R.G.
Kuller, D.R. Ostberg and F.W. Perkins;
Nering denotes Linear Algebra and Matrix Theory by E.D. Nering;
Spivak denotes Calculus by M. Spivak;
Smith denotes Mechanics by R.C. Smith and P.Smith.
Reference to Open University Courses in Mathematics take the form:
Unit Ml 00 22, Linear Algebra I
Unit MST 281 10, Taylor Approximation
Unit M201 16, Euclidean Spaces I: Inner Products
Unit M231 2, Functions and Graphs
Unit MST 282 1, Some Basic Tools.
M334 IV
INTRODUCTION
This chapter uses the differential and vector calculus developed in Sections LI to
ILL
It introduces the second major theme of this course, surfaces. The surfaces in which
we shall be interested are subsets of E 3 that locally look like some open region of
E 2 . You have already met some of these, such as planes, spheres and cylinders.
After discussing several ways of describing a surface we extend the definitions of
tangent vector, differential form and mapping to apply to them.
Finally we use differential forms to describe orientability of a surface.
M334 IV. 1
IV.l SURFACES IN E 3
Introduction
This section follows on from Sections 1.2, 1.4, 1.5, 1.7 and II. 1.
We shall be using implicitly defined curves in the plane. You first met these in
Section 1.4. Typical examples are:
¥ x
*x
(a) C^ : y  x 2 =
(b) C 2 :x 2 +y 2 =l
In general a curve is defined as a subset C: g = c, where g is a differentiable function
and c is a constant. We shall always assume that, for each point (x , y<>) in C, either
(dg/3x) (x , y ) =£ or (3g/3y) (x , y ) =£ 0, and that if C has more than one
component then we are restricting attention to just one of them. With these restric
tions the only curves that we obtain are similar to C^ and C 2 .
(a) Given a curve like C t we can parametrize it by a regular differentiable
function a from R to E 2 that is onetoone and whose inverse map a 1 from
C x to R is continuous. In our example of Cx such a parametrization is given
by a(t) = (t, t 2 ). This has inverse or 1 (x, y) = x for points in C^ .
(b) Given a curve like C 2 we can parametrize it by a periodic regular differenti
able function j3 from R to E 2 . We can also demand that if we restrict
attention to small enough open subsets of C 2 they will be the onetoone
images of small intervals of R and the inverse functions between these small
subsets will be continuous. In our example of C 2 such a parametrization is
given by j8(t) = (cos t, sin t). A typical 'local' inverse for this function is given
by IT 1 (x, y) = cos 1 x for all those points p in C 2 with y(p) > 0.
We shall need the implicit function theorem, a two dimensional form of which was
dealt with in an appendix to Unit M231, Applications of the Derivative, Section 8.
As an example of an implicitly defined curve in E 2 we can consider again the unit
circle
C: f(x, y) = x 2 + y 2 = 1
M334 IV. 1
We can use the implicit function theorem to describe the local behaviour of this
curve. The implicit function theorem states:
Theorem IV. 1. A
If the point (x , y ) belongs to C and if (3f/dy) (x , y ) =£ then there exists:
(i) a neighbourhood U of x in the real line;
(ii) a unique differentiable function g : U »R; and
(iii) a neighbourhood N of (x , yo) in the plane; such that
NnC={(t,g(t)):t€EU}.
(x »y )
*x
That is, locally the curve C can be described as the graph of a mapping g : U
Similarly we can describe C equally well in terms of y near (x , y ) if
(df/dx)(x o ,yo)*0.
►R.
8
M334 IV. 1
1 3
In our example, if x = y we can choose for U the open interval (^,4) and then g is
the function
g:tl — ►(1t 2 )*".
For a suitably chosen N we have that
NnC={(t,(l1 2 f):{<t<};
that is, N fl C is the graph of g.
In this section we shall apply an extension of the implicit function theorem to
implicitly defined subspaces of E 3 . For instance, near a point (x , y , zq) on the
unit sphere
X : f(x, y, z) = x 2 + y 2 + z 2 = 1
for which
(9f/3z) (x , y , z ) = 2z
is nonzero, the sphere can be described as the graph of a function g, where
g(x, y) = z = (l x 2  y 2 ) T .
(x ,y , (lx 2 y 2 r)
2\T\
( x o> yo»'0)
READ: Section IV.l (pages 124 131).
Comments
(i) The plane E 2 When we use E 2 , or some open set D in F^ 2 , as the domain
for a patch x we write u and v for the natural coordinate functions rather
than x and y or x l and x 2 . Consequently when we want to refer to a variable
point in E 2 we write (u, v), though strictly speaking this is the identity
function from E 2 to itself. Whichever interpretation you adopt it is unlikely
that any ambiguity will arise. When we have to use the partial derivatives of a
function f whose domain is in E 2 we write them as 8f/3u and 3f/8v.
M334IV.1
Page 124: Proper patches The conditions in the definition of a proper
patch are there to rule out certain pathological possibilities. The condition
that the mapping x is regular means that its derivative x^ is onetoone and
hence no nonzero tangent vector is mapped to a zero tangent vector. Since
derivatives preserve velocities, each curve with nonzero velocity will be
mapped to another curve with nonzero velocity. As we saw in Section 1.4
this means that x does not introduce any kinks into the patch. The image of
a patch cannot look like the following diagram.
The condition that x is onetoone rules out the possibility that the patch
might intersect itself, as in the following diagram.
The condition that x" 1 is continuous rules out the possibility that the patch
might approach itself arbitrarily closely, as in the following diagram.
10 M334 IV.l
(iii) Page 125: Definition 1.2 To prove that a subset of E 3 is not a surface
we need to find only one point that does not have any neighbourhoods that
are like smooth open discs. One way to do this is to look at the boundaries
of suitably small neighbourhoods to see if they are like circles. For instance
in the second example of Comment (ii) the boundary of a typical small
neighbourhood of a point on the line of selfintersection is shown in the
following diagram.
We shall assume that our intuition does enable us to distinguish this from a
circle.
(iv) Page 129: Curves and surfaces The curves used in both Example 1.5 and
Example 1.6 are defined implicitly. As in Section 1.4 this rules out the
possibility that they may be of any of the following forms.
(a)
(b)
(c)
This in turn ensures that the surfaces derived from them, as in Examples 1.5
and 1.6, are not pathological in any of the ways mentioned in Comment (ii).
In fact the implicit function theorem discussed in the introduction proves
that the above types of curve cannot occur. Curves (b) and (c) cannot be
expressed locally as graphs and curve (a) cannot be expressed as the graph of
a differentiable function.
When dealing with implicitly defined curves we shall always restrict ourselves
to one component at a time. In Example 1.6 this enables us to assume that
the curve is in the upper half plane. If we did not have this restriction we
would have to consider curves like the following hyperbola.
M334 IV.l
11
When this is revolved about the x axis the spaces traced out by the two
components intersect and the result is not a surface.
As mentioned in the introduction this restriction limits us to two types of
curve. The following are typical examples. We can revolve them about any
line that does not intersect them.
► x
(a) Ci : y  x
► x
For curves like C^ we shall use only onetoone parametrizations and with
curves like C 2 we shall use only periodic parametrizations. As usual all
parametrizations of implicitly defined curves will be regular.
12
M334 IV.l
Supplementary Comments
(i) Page 129: Example 1.6
ordinate fixed.
Rotating about the x axis we keep the x co
>s *( ( li> <b cos v > 92 sm v )
/(qi»q2,o)
Hence q 2 gives the radius of the circle that the point (qj , q 2 , 0) traces out as
it is revolved. When it has been revolved through an angle v its new coordi
nates are (qj , q 2 cos v, q 2 sin v).
If the point p = (pi, p 2 , p 3 ) is revolved about the x axis, its x coordinate
remains unchanged and its distance from the x axis remain unchanged. This
distance is (p 2 + p) . Hence when it has been swung round into the upper
half of the xy plane it reaches the point p = (pi , (p 2 + pi) , 0). The point p
is on the surface of revolution if and only if this point is on the original
curve. So (p! , p 2 , p 3 ) belongs to M if and only if f(p 2 , (p 2 + pi ) ) = c. That
is, M is given by M : g = c where g(x, y, z) = f(x, (y 2 + z 2 ) ).
How do we use the chain rule to prove that dg is never zero? Firstly, since g
is a mapping from E 3 to R
ag(v) = v[g] =±(g(p + tv))
t = = g *( v )
for any tangent vector v to E 3 at p. Similarly for any tangent vector w to E 2
at q then
df(w) = f*(w),
where f is a function from E 2 to R and the differential df is defined by
analogy with the definition for mappings from E 3 to R. Now g is the com
posite mapping f<>F where
F(Pi> p 2 > Pa) = (Fi (Pi» P2> Pa)> F 2 (p 1? p 2 , p 3 ))
= (Pi,(P2 2 +p!) 2 )
M334 IV.l
Hence by the chain rule
dg(v) = g*(v) = f*(F*(v)) = df(F*(v)).
Now the Jacobian matrix for F is
13
F* =
3F! 9F]
W
3F2
W
{yZTjf (y 2 +z 2 )*
Hence at points of M the mapping F* is onto, has rank 2, since either y or z
is nonzero as M does not intersect the x axis.
Assume dg is zero at p on M; then dg(v) = for all vectors v tangent to M at
p. Now if w is tangent vector to E 2 at F(p): then, since F* is onto,
w = F*(v) for some tangent vector v to E 3 at p. Hence
df(w) = df(F Jlt (v)) = dg(v) = 0,
by assumption, and so df is zero at F(p). Now for mappings from E to R
the analogue of Corollary 1.5.5 gives
At af A U 9f ^
df = ■* du +5 dv.
ou ov
Hence 8f/du and 3f/9v are both zero at F(p). But
F(p) = (Pi,(p 2 2 +P3 2 )*)
is a point of the profile curve C and so by the definition of an implicitly
defined curve 9f/3u and df/3v cannot both be zero. This contradiction
tells us that our original assumption that dg was zero at p is false; dg is never
zero on M.
Summary
Notation
x
M
M : z = f(x, y)
M : g = c
Page 124, Definition 1.1
Page 125, Definition 1.2
Page 125, line 6
Page 127, Example 1.3
Page 127, line 11
Definitions
(i) Coordinate patch x : D ►E 3
(ii) Proper patch x : D ►E 3
(iii) Surface M
(iv) Monge patch x(u, v) = (u, v, f(u, v))
(v) Simple surface M = x(D)
(vi) Graph of a function M : z = f(x, y)
(vii) Implicitly defined surface M : g = c
(viii) Cylinder
(ix) Surface of revolution
(x) Parallel
(xi) Meridian
Page 124,
Page 124,
Page 125,
Page 127,
Page 127,
Page 127,
Page 127,
Page 129,
Page 129,
Page 130,
Page 130,
Definition 1.1
line 2
Definition 1.2
line 8
line 11
Example 1.3
line 11
Example 1.5
Example 1.6
line 8
line 9
hxam
pies
(i)
Unit sphere 2
(")
Hemispherical patches on the sphere,
x(u,v) = (u,v,±(lu 2 v 2 )*)
(iii)
Sphere
(iv)
Circular cylinder,
M : x 2 + y 2 = r 2 in E 3
M334 IV. 1
Page 125, line 3
Page 126, line 7
Page 128, line 2
Page 129, line 13
Results
(i)
(iii)
(iv)
Graphs are surfaces.
An implicitly defined subset M : g = c is a
surface if the differential dg is never zero
at any point of M.
Cylinders are surfaces.
Surfaces of revolution are surfaces.
Page 127, Example 1.3
Page 127, Theorem 1.4
Page 129, Example 1.5
Page 129, Example 1.6
Techniques
(i) Recognizing a (proper) patch. Page 124, Definition 1.1 (and
Page 124, line 2)
(ii) Determining whether subsets of E 3 are surfaces by
looking for proper patches. Page 125, Definition 1.2
(iii) Recognizing that a subset of E 3 is a surface since
it is a graph, a cylinder or a surface of revolution. Page 127, Example 1.3,
Page 129, Example 1.5,
Page 129, Example 1.6
(iv) Determining whether an implicitly defined subset
of E 3 , M : g = c, is a surface by evaluating the
differential dg. Page 127, Theorem 1.4
(v) Sketching surfaces.
Exercises
Technique (i)
1. Page 132, Exercise 4.
Techniques (ii) and (iii)
2. Page 131, Exercise 1. Deleting as few points as possible from each subset
construct a surface. Explain why it is a surface.
M334 IV.l
Technique (iv)
3. Page 132, Exercise 5.
Technique (v)
4. Page 132, Exercise 6.
15
Solutions
1.
Page 132, Exercise 4.
The Jacobian matrix for the mapping x = (x x , x 2 , x 3 ) : E 2 ►E 3 is
dxi
8 Xl \
3u~
3v ]
9x 2
dx 2
9u
dv
3x 3
9x 3
(a)
(b)
The Jacobian matrix is
1 °\
u
i
This always has rank 2 and so x is regular.
Now if
x(u, v) = x(u', v')
then
(u, uv, v) = (u', u' v', v')
and so
u = u', v = v'.
Hence
x is onetoone.
Since x is onetoone and regular it is a patch
The Jacobian matrix is
J2u
ol
3u 2
o
ll
When u = this matrix has rank 1. Hence x is not regular and so is
not a patch.
16
M334IV.1
(c)
The Jacob ian matrix is
h
2u
i0 1 + 3v 2 J
and since v 2 > it has rank 2. Hence x is regular.
Suppose x(u, v) = x(u', v').
Then
(u,u 2 ,v + v 3 ) = (u',u' 2 ,v' + v' 3 ).
Hence
u = u and (v v') + (v 3  v' 3 ) = 0.
It follows that
(v v') (l + (v 2 + w' + v' 2 ))*0
and hence
(vv')(l + (v+) 2 + fv' 2 ) = 0.
Since 1 + (v + o~) 2 + 3 f v ' 2  > it follows that v = v' also and hence x is
onetoone. Since x is onetoone and regular it is a patch,
(d) If x(u, v) = x(u', v') then
cos 27ru = cos 27ru', sin 2mi = sin 27ru' and v = v'.
These conditions are satisfied whenever u = u' + n, for any integer n,
and v = v'. Hence the mapping is not onetoone and x is not a patch.
Page 131, Exercise 1.
(a) We can write M as M : g = 0, where g(x, y , z) = z 2  x 2  y 2 .
By Corollary 1.5.5 the differential of g is
dg = 2x dx 2y dy + 2z dz
and this is zero only at the origin. Hence, by Theorem IV.1.4, the
cone minus its apex is a surface.
Since the boundary of a small neighbourhood of the origin is two
circles we cannot include the origin if we require a surface.
M334IV.1 17
(b) Points not on the boundary of the closed disc have the same small
neighbourhoods as if they were considered as points of the plane
P : z = 0. Since the differential dz is nonzero it follows from
Theorem IV. 1.4 that the disc minus its boundary is a surface.
If we try to map a small open disc continuously to the closed disc in
such a way that the centre of the open disc is mapped to a point on
the boundary of the closed disc we find that we have to fold it over.
Hence the map cannot be onetoone and so the boundary points
cannot be included in the surface.
(c)
* y
The region M : xy = 0, x > 0, y > can be written as the union of two
half planes M t : x = 0, y > and M 2 : y = 0, x > 0. For any point p!
in the interior of either of these planes it is easy to find a rectangular
coordinate patch. Thus the subset of M obtained by deleting the z
axis is a surface. However, for any point p 2 on the z axis, the inter
section of these two half planes, it is impossible to find such a patch.
Any neighbourhood of p 2 has a right angle bend in it and so no
mapping x : D ►E 3 onto a neighbourhood of p 2 will be differenti
able and regular at p 2 .
18
3.
M334IV.1
Page 132, Exercise 5.
(a)
0>)
IS
By Theorem IV. 1.4, M : (x 2 + y 2 ) 2 + 3z 2 = 1 is a surface if dg
never zero on M, where g(x, y, z)  (x 2 + y 2 ) 2 + 3z 2 . Now
, 9g 3g 3g
dg = — dx + — dy + — dz, by Corollary 1.5.5.
ox dy 9z
and hence
dg = 2(x 2 + y 2 ) 2x dx + 2(x 2 + y 2 ) 2y dy + 6z dz
= 4(x 2 + y 2 ) (x dx + y dy) + 6z dz.
Hence dg is zero only at the origin and since this is not a point of M
it follows that M is a surface.
By Theorem IV. 1.4, M : z(z  2) + xy = c is a surface if dg is never
zero on M, where g(x, y, z) = z(z  2) + xy.
Now
dg = 2(z  1) dz + x dy + y dx
and hence dg is zero only at the point (0, 0, 1).
The point (0, 0, 1) belongs to M if and only if
(z(z 2) + xy)(0,0, l) = c
1(1 2) + 0.0 = c
c = l.
Hence M is a surface except possibly when c =  1.
When c = 1, the set M is given by M : z(z  2) + xy = 1 or, on
rearranging, M : (z  l) 2 + xy = 0. A sketch of the set M near the
point (0, 0, 1) shows that it is a 'double' cone. In Solution 2, above,
we saw that this is not a surface.
Page 132, Exercise 6.
M : z = y 3  3yx 2
The intersection of M with the xy plane is given by
S : = y 3  3yx 2 , z = 0.
M334 IV.l
Now if
then
19
= y 3  3yx 2
= (y 2 3x 2 )y
= (yV3x)(y+V3x)y.
The intersection consists of three lines in the xy plane. The lines are y = 0,
y  y/Sx = and y + V 3x = 0.
These lines make angles of ^ with each other.
y = V 3x
y=V3x
As a corollary to the intermediate value theorem, we can deduce that
throughout each of the six resulting regions of the plane the function f(x, y)
= y 3  3yx 2 is always positive or always negative. All we need to do is to
evaluate f at one point of each region. The values are alternately positive and
negative. Hence the surface M is alternately above and below the plane. This
surface is illustrated in Figure 5.22 on page 205.
20 M334 IV.2
IV.2 PATCH COMPUTATIONS
Introduction
This section follows on directly from Section IV. 1 and assumes familiarity with
curves, dealt with in Section 1.4, and the dot and cross product, dealt with in
Section ILL It is also useful to bear in mind the definition of the derivative of a
mapping, given in Section 1.7.
Choosing patches enables us to describe surfaces in terms of regions of the plane E 2 .
In this section we begin to see how we can use the geometry of E 2 to study surfaces.
This section contains many important examples, which will be used again and again
throughout the remainder of the course.
READ; Section IV.2 (pages 133  140).
Comment
(i)
Patches
curves
Through each point (u ,v ) in the plane E 2 there are two
ti — >(t,v )
and ti — Ku ,t),
that are parallel to the coordinate axes.
These are unitspeed curves and their velocities at the point (u , v ) are
Ui(u ,v ) and U 2 (u , v ).
ti — ►(t,v )
*u
tl ►(Uq, t)
The images of these curves under the mapping x are the u and v parameter
curves, v = v and u = u respectively. These parameter curves depend on the
choice of coordinate patch x.
Since regular mappings preserve velocity vectors, the partial velocities (the
velocities of the parameter curves) are given by
and
x u( u o» v ) = **{Ui (u , v ))
*v( u o> v o ) = x *( u 2 (u > v )).
M334 IV.2
21
Hence x u and x v are the composite mappings x 5e (Ui ) and x Jc (U 2 ). To
remind ourselves that x u and x v are functions on E 2 we sometimes write
them as x u (u, v) and x v (u, v). To calculate these explictly we look at the
parameter curves. The u parameter curve v = v is given by
a : ti — ►x(t, v ) = (x! (t, v ), x 2 (t, v ), x 3 (t, v )).
Hence
da
x u (u ,v ) = ^( u o) atx(u ,v )
/9xi , v 3x 2 . . dx 3
=( r— (u , v ), — (u , v ), — (u , v )) at x(u , v ).
ou ou ou
Since x is regular, the derivate x* is onetoone and hence it carries linearly
independent vectors into linearly independent vectors. The tangent vectors
Ui (u , v ) and U 2 (u , v ) are linearly independent in E 2 and so x u (u , v )
and x v (u , v ) are linearly independent in E 3 . In Section II. 1 we saw that
this is equivalent to the condition
x u( u o> v o) x x v (u , v ) # 0.
That is, x u X x v is never zero.
In fact this condition that x u X x v is never zero is enough to ensure that x is
regular. The condition ensures that the images of U t (u , v ) and U 2 (u , v )
under x* are linearly independent and this implies that x* is onetoone at
each point.
When O'Neill comes to calculate x u X x v he obtains the equation
X^j X Xy ~
Ui
9xj
3u~
3xj
3v~
U 2
3x 2
3u~
9x 2
9v~
U 3
8x3
3u
8x3
bv
Here L^ , U 2 , U3 are vector valued functions with domain E 3 while all the
other functions have domain E 2 and hence the equation is not technically
correct. However it is very convenient to use such expressions as long as you
are always aware of how they could be made rigorous. To make it rigorous
we could replace Ui, U 2 , U 3 by the composite functions Uj o x, U 2 ° x and
U 3 © x, or state that if the functions with the domain E 2 are evaluated at the
point (u , v ) then those with domain E 3 are evaluated at the point
p = x(u ,v ).
22
Supplementary Comments
(i) Page 136: line 1
M334 IV.2
(")
r Xj X Xy —
U t
 cos v sin u
U,
COS V COS u
sin v cos u  sin v sin u
U 3
cos v
= cos 2 v cos u Ui  cos v (cos v sin u) U 2
+((cosv sinu) (sin v sinu)  (sin v cos u)(cos v cos u)) U 3
«= cos 2 v cosu Uj * cos 2 v sinu U 2 + cos v sin v U 3
since cos 2 u + sin 2 u = 1.
Page 139: line 2, x u and x v are linearly independent
Here
x u( u » v ) = (g'( u )> h '( u ) cos v > h '( u ) sin v )
x v (u, v) = (0, h(u) sin v, h(u) cos v)
and hence
x u (u, v) X x v (u, v) = (h(u) h'(u), h(u)g'(u) cos v, h(u)g'(u) sin v).
If x u (u, v) and x v (u, v) are linearly dependent at x(u , v ) then
x u (u , v ) X x v (u , v ) = 0.
But since h(u ) > and cos v and sin v cannot simultaneously be zero, we
would have h'(u ) = g'(u ) = 0. This is impossible since by the definition (on
page 21) a parametrization is a regular curve. Hence x u (u, v) X x v (u, v) is
never zero and the partial velocities x u (u, v) and x v (u, v) are linearly
independent.
Additional Text
There are two important classes of surfaces defined in the Exercises at the end of
Section IV.2.
(i) Ruled Surfaces
READ the definition on page 140 between Exercises 3 and 4 and look at the
examples in Exercises 5 and 6.
Comment 3(u) gives the position along the base curve, 6(u) gives the 'direction' of
the line L and v gives the 'distance' along this line from the base curve.
M334 IV.2
23
x(u, v)
origin
5(u) need not be the same length and direction for each u.
(ii) Quadric surfaces
READ the definition on page 142, between Exercises 9 and 10, and browse
through the examples in Exercises 10 to 11.
Summary
Notation
x u (u ,v ) and x v (u ,v )
x u and x v
Page 134, Definition 2.1
Page 134, line 12
Definitions
(i) u and v parameter curves, v = v and u = u
(ii) Partial velocities, x u (u ,v ) and x v (u ,v )
(iii) Parametrization
(iv) Crosssectional curve of a cylinder
(v) Rulings of a cylinder
(vi) Ruled surface
(vii) Rulings
(viii) Ruled form
(ix) Base curve
(x) Director curve
(xi) Quadric surface
Page 133, line 9
Page 134, Definition 2.1
Page 135, Definition 2.3
Page 138, line 9
Page 138, line 11
► Page 140, lines  16 to  8
Page 142, line 16
24 M334 IV.2
Examples
(i) The geographical patch on the sphere:
x(u, v) = (r cos v cos u, r cos v sin u, r sin v),
where
uG (7T, 7r), vG (£, «) Page 135, line 3
(ii) Parametrization of a cylinder:
x(u, v) = («! (u), ol 2 (u), v) Page 137, Example 2.4
(iii) Parametrization of a surface of revolution:
x(u, v) = (g(u), h(u) cos v, h(u) sin v) Page 138, Example 2.5
(iv) Parametrization of a torus of revolution:
x(u, v) = ((R + r cos u) cos v,
(R + r cos u) sin v, r sin u) Page 139, Example 2.6
(v) Parametrization of a ruled surface in ruled form:
x(u, v) = j8(u) + v5(u) Page 140, line  12
(vi) Helicoid:
x(u, v) = (u cos v, u sin v, bv) Page 141, Exercise 7
Results
(i) x : D ►E 3 is regular if and only if x u X x v is
never zero. Page 136, line 6
(ii) The partial velocities are the images of the standard
basis vectors under the derivative mapping x^:
x*(Ui ) = x u , x„ c (U 2 ) = x v . Text, page 20
Techniques
(i) Description of the parameter curves. Page 133, line 9
(ii) Evaluation of partial velocities. Page 134, line 7
(iii) Testing regularity of a parametrization. Page 136, line 6
(iv) Use of standard parametrization of cylinders and
surfaces of revolution. Page 137, Example 2.4,
Page 138, Example 2.5
(v) Finding the ruled form of a surface, in certain
simple examples. Page 140, line  16
(vi) Familiarity with the shapes and equations of
quadric surfaces. Page 142, line 16
M334 IV.2
Exercises
25
Techniques (i), (ii) and (Hi)
1. Page 141, Exercise 7.
Technique (Hi)
2. Page 140, Exercise 2.
Technique (iv)
3. Page 141, Exercise 6.
4. Page 140, Exercise 1.
Technique (v)
5. Page 140, Exercise 4.
Technique (vi)
6. Page 142, Exercises 10 and 11. Only sketch the surfaces.
7. Page 142, Exercise 12(a) and (b).
Solutions
1. Page 141, Exercise 7.
(a)
(b)
To check that x is a patch we must show that it is onetoone and
regular. If x(u, v) = x(u', v') then
u cos v = u' cos v', u sin v = u' sin v' and v = v\
Since v = v' and the cosine and sine functions cannot be simul
taneously zero it follows that u = u'. Hence x is onetoone.
Now
Xy X Xy —
U,
cos V
u sin v
U,
sin v
u cos v
U 3
b
= (b sin v,  b cos v, u).
Hence x u X x v  2 = b 2 + u 2 > 0, and so x u X x v i= 0. Thus x is
regular and therefore a patch.
The u parameter curve v = v is
1 1 — >x(t, v ) = (t cos v , t sin v , bv )
= (0, 0, bv ) + t(cos v , sin v , 0),
which is a straight line parallel to the line in the xy plane making an
angle v with the x axis.
26
M334 IV.2
The vparameter curve u = u is
ti — »"x(u , t) = (u cos t, u sin t, bt).
As t increases by 2flr the curve goes once round the z axis, in a circle
of radius u , and rises 27rb. The vparameter curves are helices about
the z axis.
(c) In order to express the helicoid in implicit form we must find some
function g and constant c such that g(p) = c if and only if p =
x(u , v ) for some (u , v ).
If p = x(u ,v )then
and
p 3 =bv , v =p 3 /b
Pi =u cos(p 3 /b), p 2 =u sin(p 3 /b).
M334 IV.2 27
Hence for each point p of the helicoid
Pi sin (p 3 /b) = p 2 cos (p 3 /b)
so if a point belongs to the helicoid it also belongs to the implicitly
defined surface
M : x sin(z/b)  y cos(z/b) = 0.
We need to show conversely that any point of M belongs to the
helicoid. Suppose p satisfies the above relationship. Then since b #
we can set v = p 3 /b and we obtain
p! sin v =P2 cos v .
For any value of v at least one of cos v , sin v must be nonzero.
Suppose cosvq^O. Then we can find u =pi/cosv . Now
p 2 cos v = Pi sin v = u cos v sin v and so, since cos v =£ 0,
p 2 =u sin v . Hence we can write p in the form (u cos v , u sin v ,
bv ) and so it belongs to the helicoid.
A similar argument works when sin v =£ 0.
Hence the helicoid is the surface
M : x sin(z/b)  y cos(z/b) = 0.
2. Page 140, Exercise 2.
By Lemma II. 1.8,
 x u X x v  2 = (x u • x u ) (x v • x v )  (x u • x v ) 2
*=EG  F 2 .
Now x is regular if and only if x u X x v =£ 0. But x u X x v is never zero if and
only if x u X Xyfl is never zero, which is true if and only if EG  F 2 is never
zero.
3. Page 141, Exercise 6.
x(u, v) = 0(u) + vq.
Hence x u = 3'
and x v = q:
x u X u v = 0' X q,
and so x is regular if and only if 3' X q is never zero.
In Example 2.4 the curve is in the xy plane and q = (0, 0, 1).
Page 140, Exercise 1.
(a) A parametrization of the profile curve is ui — ►(u, cosh u, 0).
A parametrization of the catenoid is
x : (u, v)i — ►(u, cosh u cos v, cosh u sin v)
since rotation about the x axis keeps the x coordinate fixed.
28
M334 IV.2
u, cosh u, 0)
(b)
x(u, v) = (u, cosh u cos v, cosh u sin v)
C : (z  2) 2 + y 2 = 1 is a circle in the yz plane of radius one and
centre (0, 2). A possible parametrization is
ui — ►(O, sin u, 2 + cos u).
A parametrization of the torus is
x : (u, v)i — >((2 + cos u) sin v, sin u, (2 + cos u) cos v),
since rotation about the y axis keeps the y coordinate fixed.
y
((2 + cos u) sin v, sin u, (2 + cos u) cos v)
, sin u, 2 + cos u)
>z
(c) We cannot follow the same procedure as in parts (a) and (b) since the
profile curve cuts the axis of rotation and hence the standard
'parametrization' will not be regular. However we can describe the
surface in terms of a Monge patch.
M334 IV.2
29
P = (Pl»P2»P3)
A point is on the paraboloid M if the square of its distance from the
z axis is equal to its height above the xy plane. Thus p = (p x , p 2 , P3 )
belongs to M if and only if pj + p\ = p 3 . Hence M : z = x 2 + y 2 and
we can parametrize it by mapping each point in the xy plane to the
unique point on the surface above it. This gives the single patch
x : (u, v)i — >>(u, v, u 2 + v 2 ).
Page 140, Exercise 4.
Look at the intersection of M with the plane x = c, for some constant c. This
is the straight line z = cy in the plane x = c. It passes through the points
(c, 0, 0) and (c, 1, c). We can write it as
v.— *(c,0,0) + v(0, l,c).
Letting the point (c, 0, 0) vary along the x axis we obtain the para
metrization
(u,v)i — >(u, 0, 0) + v(0, l,u),
which is in ruled form.
Similarly looking at the intersection of M with planes of the form y = c we
find the parametrization
(u,v)i— ►(0,v,0)+u(l > 0,v).
Alternatively, since M : z = xy, M is described by the Monge patch
x : (u, v)i ►(u, v, uv),
from which the above two ruled forms follow.
30
6. Page 142, Exercises 10 and 11.
10(a). Ellipsoid
M334 IV.2
A z
10(b). Elliptic hyperboloid
M334 IV.2
10(c). Elliptic hyperboloid (two sheets)
31
1 1 . Elliptic parab oloid
32
7.
M334 IV.2
Page 142, Exercise 12
(a)
(b)
The mapping x : E 2 ► E 3 does map into M since
_ (a(u + v)) 2 (b(uv)) 2
a 2 b 2 *
We need to show further that it maps onto M. For any point p
belonging to M we need to solve the equations
Pi = a(u + v )
p 2 =b(u  v )
p 3 =4u v .
We can solve the first two, obtaining
_Pi , P2 _Pl P2
Uo = 2a" 2b' Vo= 2a~2b
and since p belongs to M the third equation is also satisfied.
Next we need to check that x is onetoone. Suppose
x(u, v) =x(u', v'):
then
U + V = U + V
I I
u  v = u  V
and hence u = u', v = v\ The mapping x is onetoone.
Now we need to check that x is regular: we use the usual method.
x u x x u ~
Ui
u 2
b
U 3
4v
a b 4u
= (4b(u + v), 4a(v  u),  2ab).
Now, for the definition of the hyperbolic paraboloid to be meaning
ful we need a,b > 0. Hence 2ab ¥= and so x u X x v is never zero.
This in turn proves that x u and x v are linearly independent and
hence x is regular.
Thus x is indeed a patch and to complete the proof we need only
prove that x is proper. To prove that x is proper we have to show
that the inverse function x  l
given by the formula
Mi ►E 2 is continuous. Since x 1 is
JL + JL
2a 2b
x
2a"
_y_
2b
continuity follows from standard results in analysis.
We can write x in ruled form as follows
(i) x(u, v) = (au, bu, 0)+v(a,b,4u)
and
(ii) x (u, v) = (av, bv, 0) + u(a, b, 4v).
M334 IV.3 33
IV.3 DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS
Introduction
This section follows on from the previous section, IV.2, and follows the develop
ment of Chapter I, Sections 1 to 4. You will also need the chain rule dealt with in
Section 1.7.
The section begins with certain rather technical but intuitively obvious definitions
and results. Mappings that you would expect to be differentiable turn out to be
differentiable, and, given a coordinate patch, we can use it to give explicit descrip
tions of curves in a surface. Do not worry about the technicalities.
We then show that the partial velocities, for a given patch, do tell us something
independent of the choice of patch. The tangents to any curve in the surface can be
written as a linear combination of them. These tangents form the tangent planes to
the surface. We use them to define a directional derivative for functions defined only
on the surface, and not on the whole of E 3 .
READ: Section IV.3 (pages 143149). Do not spend much time on pages
143  145. Just try to understand the results.
Comments
(i) Differentiation on M and on E 3 We have now met several different defini
tions of differentiation with domain or codomain either E 3 or some surface
M. If we are dealing with a function F : E n ►E 111 between Euclidean spaces
and we are using the definition of differentiability given in Section 1.7 we
say that F is Euclidean differentiable. If we are using either of the definitions
of this section for functions with domain or codomain some surface M we
indicate it by saying that G : M >E n is differentiable or F : E n »M is
differentiable. We need to compare these definitions of differentiability with
the Euclidean definition.
Directly from the definitions we have that G : M ►E 11 or F : E n ►M are
differentiable if their respective coordinate expressions are Euclidean
differentiable.
Suppose we have a function F : E n ►M. Then we can also consider it as a
function F : E n ►E 3 . If we know that F is differentiable as a function
from E n to M then we would hope that F is also Euclidean differentiable. It
is. Since F : E n  — >M is differentiable it follows, by definition, that
x _1 o F : E n ►E 2 is Euclidean differentiable, for each patch x. But by the
definition of a patch, each patch x : D ►E 3 is Euclidean differentiable.
Consequently the composite mappings x o x _1 o F : E n >E 3 are Euclidean
differentiable. But each of these mappings is just the restriction of F to those
points whose image under F is in x(D). Since the patches cover M we can
choose enough "x"s to deal with F at every point. Hence F : E n >E 3 is
Euclidean differentiable.
Theorem 3.2 gives us the converse of this result. If F : E n ►E 3 is a
Euclidean differentiable mapping whose image lies in M then it is also
differentiable considered as a function F : E n ►M.
34 M334 IV.3
The corollaries are proved as follows.
Corollary 3.3 Since y is a patch it is a differentiable function y : D ►E 3
whose image lies in M. Hence, by Theorem 3.2, it is also differentiable
considered as a function y : D >M. By definition this just means that
x _1 o y is a Euclidean differentiable mapping between suitable subsets of E 2 .
Corollary 3.4 Since x 1 y is Euclidean differentiable
x _1 y(u, v) = (u(u, v), v(u, v)),
where the Euclidean coordinate functions u, v are differentiable. Applying
the mapping x to each side gives the result. The mapping x~ 1 y = (u, v) gives
us a coordinate transformation.
Theorem 3.2 can also be used to give a converse of Lemma 3.1. Suppose a
function o: : I »M is such that
a(t) = x( ai (t), a 2 (t))
for differentiable functions aj and a 2 . Then the composite x(a 1 ,a 2 ) is a
Euclidean differentiable mapping into E 3 , whose image lies in M. Hence by
Theorem 3.2 its restriction a : I ►M is differentiable; that is, a is a curve.
As a corollary to this we find that the parameter 'curves' are in fact curves in
M since they can be written as
ti >x(t, v )
ti — ►x(u ,t),
where the functions given by t< ►t, ti >v , ti — ►uq are all differentiable.
When no confusion can arise we omit the word 'Euclidean' from the
expression 'Euclidean differentiable'.
(ii) Page 147: line 1 We have two mappings F and x with the following
Euclidean coordinate functions:
F=(a!,a 2 ) and x = (x 2 , x 2 , x 3 ).
The composite mapping x o F is a. The composite (chain) rule gives
a* = x*(F) o F*.
In terms of Jacobian matrices this becomes
<\
«2
% 't
3u~
9x2
3u~
3u
3x 2
9x 3
3v
ai
32
F(t)
Multiplying out we obtain
a'(t) = x u (F(t))a;(t) + x v (F(t))a 2 (t)
= x u ( ai (t), a 2 (t))aj (t) + x v ( ai (t), a 2 (t))ai (t).
M334IV.3
(iii)
35
Page 147: vector fields Since x u (u ,v ) and x v (u ,v ) are tangent
vectors at x(u , v ) we can cons